Circular Motion in Horizontal and Vertical Planes Physics Notes

Circular Motion in Horizontal and Vertical Planes Physics Notes
1. Motion in a Horizontal Plane:
The figure shows a mass m tied to an end of a string of length L. The mass m moves in a horizontal plane with a constant speed. As the mass moves in the circle, the string sweeps a cone of an angle θ with the surface where θ is the angle made by the string with the normal. The forces that act on the mass m at any instant are shown in the figure. If T is the tension in the string, then the components of T will be T cosθ and T sinθ. There is no vertical acceleration on m. So, the component T cos θ balances the weight W of m. This way

Circular Motion in a Horizontal Plane

Here, r is the radius of the circular motion.

If the length of the string is L, then from the figure.
r = L sinθ
and v = rω = r\(\left(\frac{2 \pi}{\tau}\right)=\frac{2 \pi r}{\tau}=\frac{2 \pi L \sin \theta}{\tau}\)
where, τ is the time period.
Putting these in equation (3), we get

Here, the value of θ cannot be 90°, because then x will be zero or v = ∞.
The maximum value of x will be
τ_max = 2π\(\sqrt{\frac{L}{g}}\)

It is possible for a very small angle (θ ≈ 0°) so that cos θ ≈ cos 0° = 1.
Equation (5) represents the formula for the time period of a simple pendulum. With this similarity, the above device is called the conical pendulum.

2. Motion in a Vertical Plane:
Consider a body of mass m tied at the end of a string and whirled in a vertical circle of radius r. Let v₁ & v₂ be velocities of the body and T₁ and T₂ be tensions in the string at the lowest point A and the highest point B respectively. The velocities of the body at points A and B will be directed along tangents to the circular path at these points while tensions in the string will always act towards the fixed point O as shown in the figure. At the lowest point A, a part of tension T₁ balances the weight of the body and the remaining part provides the necessary centripetal force. Therefore.
T₁ – mg = \(\frac{m v_{1}^{2}}{r}\) …(1)
At the highest point, the tension in the string and the weight of the body together provide the necessary centripetal force.

T₂ + mg = \(\frac{m v_{2}^{2}}{r}\) …(2)
Let us now find the out minimum velocity, the body should possess at the lowest point so that the string does not slack when it is at the highest point. The body is then said to just loop the vertical circle. It is obvious that the velocity at the lowest point will be minimum. When the velocity at the highest point is also minimum.

Minimum velocity at the highest point:
From equation (2), it follows that the velocity at the highest point will be minimum when the tension in the string at the highest point is zero.
T₂ = 0 …(3)

In that case, the whole of the centripetal force will be provided by the weight of body. Therefore in such a case, the equation (2) becomes:
0 + mg = \(\frac{m v_{2}^{2}}{r}\)
or v₂ = \(\sqrt{g r}\) …(4)

This is the minimum velocity the body should possess at the top so that it can just loop the vertical circle without the slackening of the string. In case the velocity of the body at point B is less than \(\sqrt{g r}\). The string will slack and the body will not loop the circle. Therefore, a body will just loop the vertical circle if it possesses velocity equal to \(\sqrt{g r}\) at the top.

Minimum velocity at the lowest point: According to the principle of conservation of energy.
K.E. of the body at point A = (P.E. + K.E) of the body at point B

As said earlier, when the velocity at the highest point is minimum. The velocity at the lowest point will also be minimum.

Setting v₂ = \(\sqrt{g r}\), we have
v²₁ = 4gr + gr
or v₁ = \(\sqrt{5 g r}\) …(5)

’The equation (5) gives the magnitude of the velocity at the lowest point with which body can safely go round the vertical circle of radius r or can loop the circle of radius r. Let us find out the tension in the string at the lowest point in such a case.

In equation (1),

Centripetal Force: When a rigid body moves along a circular path with uniform speed, its direction is changed continuously. Due to inertia, at every point of the circular path; the body tends to move along the tangent to the circular path at that point. To move a body along a circular path; an external force is required, which will deflect the body from its straight path to the circular path at every point of the path.

“An external force required to make a body move along a circular path with uniform speed is called the centripetal force”.

A body in uniform circular motion is in a continuously accelerated motion and the acceleration is directed along the radius of the circular path and towards the center. It is called centripetal acceleration and as proved in the last section, it is given by
a_c = \(\frac{v^{2}}{r}\)
Where r is the radius of circular path and v, the uniform linear speed of the body. Therefore, the centripetal force required to move a body of mass ‘m’ with uniform speed v along a circular path of radius r is given by
F_c = ma_c = \(\frac{m v^{2}}{r}\) …(1)
Since v = rω, the above equation may be written as
F = mω²r …(2)

Examples centripetal force:

• For an electron revolving around the nucleus the centripetal force provided by the electron static attraction between the electron and the nucleus.
• For a stone rotated in a circle the tension in the string provides the centripetal force.
• For a car taking circular turns on a horizontal road, the centripetal force is provided by the force of friction between the tires and the road.

Physics Notes

Centripetal Acceleration Physics Notes

Centripetal Acceleration:
We will consider the circular motion of a particle as shown in the figure. At any instant t, the particle is at P, where its angular displacement is θ.

Now, we will draw a unit vector PA = â_r on point P, in the outward direction from the radius r of the circle, and another unit vector PB = â_t, tangent to the point and in the direction of the angle θ as shown in the figure.

We call â_r as radial unit vector and â_t as a tangential unit vector.

Now, draw a line PX’ parallel to X-axis and PY’ parallel to Y-axis, as depicted in the figure.

Calculation of Acceleration in Circular Motion

According to the figure,
\(\frac{\overrightarrow{P A}}{|\overrightarrow{P A}|}\) = î cos θ + ĵ sin θ
â_r = î cos θ + ĵ sin θ ….(1)

Here, | \(\vec{PA}\) | = 1 and î and ĵ are the unit vectors along X-axis and Y-axis.
Similarly,
\(\frac{\overrightarrow{P B}}{|\overrightarrow{P B}|}\) = -î sin θ + ĵ cos θ
or
â_t =-î sinθ +ĵ cosθ …(2)
Now, the position vector of the particle at time 2
will be
\(\vec{r}=\overrightarrow{O P}\) = OP âr
or, \(\vec{r}\) = r (î cos θ + ĵ sin θ) …(3)
Now, taking the derivative of equation (3) w.r.t. t, we get the velocity of the particle.
So, the velocity of the particle

The above equation shows that the term rω is the speed of the particle at time t [: v = rω] and its direction is toward â_t or tangential.

We can find out the acceleration of the particle at any instant of time t by taking the derivative of equation (4).
∴ The acceleration of the particle,

and equations (1) and (2).

In equation (6) we observe that the acceleration of particle a has two components.
(i) \(\vec{a}\) r = -ω râ_r directed along -â_r or towards the centre of the circle.
That is why it is called centripetal acceleration.

It is directed along the tangent of the point. Its value is rex. For a non-uniform circular motion of a particle, there will be these two accelerations, but for the uniform circular motion of a particle, v is definite and so is co
as (v = rω). Therefore, \(\frac{d v}{d t}\) = r\(\frac{d \omega}{d t}\) = 0. Hence, the value of tangential acceleration at will be zero. So, the centripetal acceleration (from equation (6)).
\(\overrightarrow{a_{r}}=\vec{a}\) = -ω2 râ_r ………..(7)

This acceleration will be directed towards the centre of the circle and the magnitude of the centripetal acceleration will be
â_r = ω²r = \(\frac{v^{2}}{r^{2}} r=\frac{v^{2}}{r}\) …….(8)

Here, we should note that u is constant but the direction of the particle keeps on changing. So, there is a change in velocity of the particle and hence, acceleration is produced.

The net acceleration in a circular motion :

Centripetal force: A force required to make a body move along a circular path with uniform speed is called centripetal force. This force is always acting along the radius and towards the center of the circular path.

Expression for centripetal force: We drive the centripetal acceleration in article number 4. 10.

Physics Notes

Circular Motion Physics Notes

Circular Motion: When a body moves such that it always remains at a fixed distance from a fixed point, then its motion is said to be circular motion. The fixed distance is called the radius of the circular path and the fixed point is called the center of the circular path.

Uniform Circular Motion: Circular motion performed with a constant, speed is known as uniform circular motion.

Angular Displacement: Angle swept by the radius vector of a particle moving on a circular path is known as angular displacement of the particle.

Let the angular position of the particle at times t₁ and t₂ are θ₁ and θ₂ respectively. Then angular displacement.
Δθ = θ₂ – θ₁

Note:

• Angular displacement is a vector quantity.
• Its direction is perpendicular to the plane of rotation and given by the right-hand screw rule.
• Clockwise angular displacement is taken as negative and anticlockwise displacement as positive.
• Its unit is radian (in M.K.S.).
• Always change degree into radian. If it occurs in numerical problems, 1 radian = \(\frac{180^{\circ}}{\pi}\)
• It is a dimensionless quantity i.e., [M°L°T°].

Relation between angular displacement and linear displacement:
Since, angle = \(\frac{\text { Arc }}{\text { Radius }}\)
∴ Angular displacement = \(\frac{\text { Arc } P P^{\prime}}{\text { Radius }}\)
Δθ = \(\frac{\Delta S}{r}\)
∴ For circular motion
Δs = rΔθ

Angular Velocity: The rate of change of angular displacement of a body with respect to time is known as angular velocity. It is represented by ω (omega).

• It is a vector quantity.
• Its direction is the same as that of angular displacement i.e., perpendicular to the plane of rotation.
• If the particle is revolving in the clockwise direction then the direction of angular velocity is perpendicular to the plane downwards. Whereas in the case of the anticlockwise direction the direction will be upwards.
• Its unit is radian/sec.
• Its dimension is [M°L°T^-1].

Average Angular Velocity: It is defined as the ratio of total angular displacement to total time.
\(\vec{\omega}_{a v}=\frac{\text { Totalangulardisplacement }}{\text { Total timetaken }}\)
or \(\vec{\omega}_{a v}=\frac{\Delta \theta}{\Delta t}\)

Instantaneous Angular Velocity: Angular velocity of a body at some particular instant of time is known as instantaneous angular velocity.
or
Average angular velocity evaluated for a very short duration of time is known as instantaneous angular velocity.

Note: Instantaneous angular velocity can also be called simply angular as velocity.

Relation between linear velocity and angular velocity
We know that angular velocity

The period of uniform circular motion:
Total time taken by the particle performing uniform circular motion to complete one full circular path is known as time period.

In one time period total angle rotated by the particle is 2π and time period is T.
Hence, angular velocity
ω = \(\frac{2 \pi}{T}\)
or
T = \(\frac{2 \pi}{\omega}\)

Frequency: Number of revolutions made by the particle moving on a circular path in one second is
known as frequency.
f = \(\frac{1}{T}=\frac{\omega}{2 \pi}\)
⇒ ω = 2πf
Angular Acceleration: The rate of change of angular velocity is defined as angular acceleration
If Δω be the change in angular velocity in time Δt, then angular acceleration

• It is a vector quantity.
• Its direction is the same as that of change in angular velocity.
• Its unit is rad/sec.
• Dimension: [M°L°T^-2].

Relation between angular acceleration and linear acceleration
We know that
Linear acceleration = Rate of change of linear velocity
⇒ a = \(\frac{d v}{d t}\) …..(i)
Angular acceleration = Rate of change of angular velocity
⇒ a = \(\frac{d \omega}{d t}\) …….(ii)
From equation (i) and (ii)

Equation of Linear Motion and Rotational Motion

Physics Notes

Advantages and Disadvantages of friction Physics Notes

Friction as a necessity (Advantages of friction):
1. Friction help us to walk. When we press the ground with our foot backward, we receive, forward reaction due to friction between the foot and the ground. Had there been no friction, the foot will simply slip away.

2. Friction helps us to write on a blackboard or on paper. As we write, the chalk particles stay on the board due to friction.

3. Friction is necessary for the movement of vehicles on the road. It is the static frictional force which makes the acceleration and retardation of vehicles possible on the road.

4. Friction is helpful in tying knots in the rope and strings.

5. Brakes make use of friction to stop the vehicles. Special high friction materials are used for the brakes of automobiles.

6. Treading of tires of vehicles is done in order to increase friction.

7. The transfer of motion from one part of a machine to the other part through belts and pulleys will not be possible without friction.

8. Cleaning with sandpaper will not be possible without friction.

9. Adhesives will lose their purpose.

Disadvantages of Friction:

• Friction always opposes the relative motion between any two bodies in contact. Therefore, extra energy has to be spent on overcoming friction. This reduces the efficiency of a machine.
• Friction causes wear and tear of the parts of machinery in contact. Thus, their lifetime reduces.
• Frictional forces result in the production of heat, which causes damage to the machinery.

Methods of Reducing Friction:
We can reduce friction:
1. By polishing: Polishing makes the surface smooth by filling space between the depressions and projections present in the surface of the bodies at the microscopic level. Thus interlocking between the body and the surface. Therefore friction reduces.

2. By lubricating: When oil or grease is placed between the two surfaces in contact, it prevents the surface from coming in actual (direct) contact with each other. This converts solid friction into liquid friction which is very small in magnitude. The lubrication reduces friction as well as reduces the heating of moving parts.

3. By proper selection of material: Since friction depends upon the nature of the material used hence it can be largely reduced by proper selection of the material.

4. By converting sliding friction into rolling friction: The friction during rolling becomes much smaller than during sliding of all the frictional forces. The rolling frictional force is minimum. Hence, in order to avoid the wear and tear of machinery, it is required to convert kinetic frictional force into rolling frictional force. The wheels and ball bearings are used to reduce friction as they convert sliding friction into rolling friction.

5. By streamlining: It implies giving a typical shape (sharp in the front) to high-speed vehicles. The fluid friction decreases due to streamlining. It. is for this reason that airplanes, jets, etc are given such typical shapes at the front.

Physics Notes

Friction Physics Notes

Friction:
If we slide or try to slide a body over a surface, the motion is resisted by a bonding between the body and the surface. This resistance is represented by a single force and is called friction force.

The force of friction is parallel to the surface and opposite to the direction of intended motion.
Look at this diagram, at first, the wooden block is at rest. A push causes the block to slide across the desk. The force of the push (big F) keeps the block moving. To stop a moving object, a force must act in the opposite direction to the direction of motion. As the block slides across the desk, it slows down and stops moving. The force that opposes the motion of an object is called friction.

• The force of friction is parallel to the surface and opposite to the direction of intended motion.
• Frictional force refers to the force generated by two surfaces that contact and slides against each other. These forces are mainly affected by the surface texture and amount of force impelling them together. The angle and position of the object affect the amount of frictional force.
• If an object is placed against an object, then the frictional force will be equal to the weight of the object.
• If an object is pushed against the surface, then the frictional force will increase and becomes more than the weight of the object.
• The maximum amount of friction force that a surface can apply upon an object can be easily calculated with the use of the given formula.
  Friction force = µ normal
  f = µN
• According to the old view, the roughness of surfaces is the cause of friction. A surface that appears very smooth to the naked eye is found to have irregularities (roughness). When two bodies are in contact with each other, the irregularities in the surface of one body get interlocked in the irregularities of. the other surface.
• The modern view is that it añses on account of strong atomic or molecular forces of attraction between the two surfaces at the point of actual contact.

Types of Friction:
Frictional forces are of three types:

1. Static frictional force
2. Kinetic frictional force
3. Rolling frictional force

 

1. Static Frictional Force: Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object.

Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration.
a = \(\frac{F}{m}\)

Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs.

Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.

1. The static friction depends upon the nature of the surfaces of the two bodies in contact.
2. The static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play only when the applied force tends to move the body.
3. The static friction is a self-adjusting force.

Limiting Friction: If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move is called limiting friction.

1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them.
(f_s)_max ∝ N
(f_s)_max = µ_sN ……..(1)
Where the constant of proportionality µs is called the coefficient of static friction. Its value depends upon the nature of the surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non-polished. For example, when two polished metal surfaces are in contact, µ_s = 0.2, when these surfaces are lubricated, µ_s = 0.1

The value of µs lies between 0 and 1 or 0 < µ_s < 1
∵ N = Mg
∵ (F_s)max = µ_smg ……….(2)

2. Direction of the force of limiting friction is always opposite to the direction in which one body is on the verge of moving over the other.

3. Coefficient of Static Friction: µs is called the coefficient of static friction and is defined as the ratio of the force of limiting friction and normal reaction
From(1), µ_s = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M⁰L⁰T⁰]
Unit: It has no unit.
The value of µs does not depend upon the apparent area of contact.

2. Kinetic Friction: We know that when the applied force on a body is small, it may not move but as the applied force becomes greater than the force of limiting friction, the body is set into motion. The force of friction acting between the two surfaces in contact which are moving relatively, so as to oppose their motion, is known as kinetic frictional force.

1. Kinetic friction is directly proportional to the normal reaction i. e.,
f_k ∝ N
or f_k = µ_kN………(1)
Where µk, is called the coefficient of kinetic friction.
∵ N = mg
∴ f_k = µ_kmg ……(2)

2. Value of µ_k depends upon the nature of the surface in contact.

3. Kinetic friction is always lesser than the limiting friction
f_k < (f_s)_max
∴ µ_k < µ_s
i. e., the Coefficient of kinetic friction is always less than the coefficient of static friction. Thus we require more force to start a motion than to maintain it against friction. This is because once the motion starts actually; the inertia of rest has been overcome. Also when motion has actually started, irregularities of one surface have little time to get locked again into the irregularities of the other surface.

4. Kinetic friction does not depend upon the velocity of the body.
The graph between applied force and force of friction. If we plot a graph between the applied force and the force of friction, we get the curve of the type shown in the figure. The part OA represents static friction, fs which goes on the increase with the applied force. The body remains at rest till the applied force does not exceed OK and at A, the static friction is maximum. This represents the limiting friction (F_s)_max (= AK or OM). Once the body starts moving, the force of friction drops to a value CL or ON, slightly less than limiting static friction. Thus CL represents the kinetic friction (F_k)(= ON).

As the portion BC of the curve is parallel to the X-axis, therefore, kinetic friction does not change with the applied force. It remains constant, whatever be the applied force.

Further, dynamic friction of friction may be of two types:

1. Sliding friction
2. Rolling friction

1. Sliding Friction: The opposing force that comes into play when one body is actually sliding over the surface of the other body, is called sliding friction. For example, a flat block is moving over a horizontal table.

2. Rolling Friction: Frictional force which opposes the rolling of bodies (like a cylinder, sphere, ring, disc) over any surface is called rolling frictional force.
Rolling friction is directly proportional to the normal reaction (N) and inversely proportional to the radius (r) of the rolling cylinder or wheel.
∴ Frolling = µ_r \(\frac{N}{r}\)
Here μ_r is called the coefficient of rolling friction.

Rolling friction is often quite small as compared to sliding friction. That is why heavy loads are transported by placing them on carts with wheels.

Cause of Rolling Friction: When anybody rolls over any surface it causes a little depression and a small hump is created just ahead of it. The hump offers resistance to the motion of the rolling body, this resistance is rolling frictional force. Due to this reason only, hard surfaces like cemented floors offer less resistance as compared to the soft sandy floors because the hump created on a hard floor is much smaller as compared to the soft floor.

Friction is a necessary evil: Although frictional force is a non-conservative force which opposes the motion. It always causes lots of wastage of energy in the form of heat yet it is very useful to us in many ways. That is why it is considered as a necessary evil.

Physics Notes

Systems With Variable Mass Physics Notes

Systems With Variable Mass:
Usually, we assume that the total mass of a system remains constant but in some incidents mass changes in which the study of the motion of the system can be done by using the laws of motion.

According to the second law of Newton

This equation shows that the mass of the system changes with time. In the form of variable mass systems, conveyor belt loading arrangements and rocket propulsion are the main examples.

1. During rain, when raindrops fall downwards then due to evaporation or condensing of moisture at the surface of the drop, may change the mass of the drop.

Let the mass of the raindrop be m, falling with the velocity \(\vec{v}\). Let the rate at which it gets condensed by \(\frac{d m}{d t}\) when it moves with velocity v₀ at the surface of raindrop. Then the total rate of change in momentum of the drop = change in momentum due to acceleration + Change in momentum due to condensation of the moisture

Equation (1) can be solved in different situations

2. Conveyor Belt Loading System: This arrangement is an example of variable mass system. As shown in figure. It allows to fall metallic mixture from the Hoffer at a study rate.

Conveyer belt moves forward with constant velocity by the help of carrier belt roller. Necessary force needed to reach the metallic mixture to proper place, is provided by the motor which is attached to the roller.

Let us assume that a metallic mixture falls at a study rate \(\frac{d m}{d t}\) on the conveyor belt. At any instant, the dt mass present at the belt is \(\vec{v}\) and mass of the belt is M.

If the constant velocity of conveyor belt is v then the total momentum of the system.
\(\vec{p}\) =(m + M) \(\vec{v}\) ……(i)
∴ The necessary force to move the belt
\(\vec{F}=\frac{d \vec{p}}{d t}\) (by the second law of motion) dt

Since the mass M and velocity \(\vec{v}\) of the belt is constant.

The above expression shows that the necessary external force depends upon the change in mass. This force does not depend on the mass of the belt.

3. Motion of a Rocket: As the fuel in the rocket is burnt and the exhaust gas is expelled out from the rear of the rocket in the downward direction. The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expel it. This force exerted by the exhaust gas on the rocket provides upthrust to the rocket. The more gas is ejected from the rocket, the mass of the rocket decreases.

To analyse this process let us consider a rocket being fired in an upward direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to the gravity with height.

Figure (a) shows a rocket of mass ‘m’ at a time ‘t’ after its take-off moving with velocity y. Thus at time ‘t’ momentum of the rocket is equal to Mv
Thus, \(\vec{p}_{c}=M \vec{v}\) …….(1)
Now after a short interval of time dt, gas of total mass dm is ejected from the rocket. If Vg represents the downwards speed of the gas relative to the rocket then the velocity of the gas relative to the earth is
\(\overrightarrow{v_{g e}}=\vec{v}-\overrightarrow{v_{g}}\)

At time (t + dt), the rocket and unburned fuel gases mass (M – dm) and it moves with the speed of \((\vec{V}+d \vec{V})\). Thus, the momentum of the rocket is:
= (M – dm)\((\vec{v}+d \vec{v})\)

Total momentum of the system at time (t + dt) is
\(\) = dm\(\left(\vec{v}-\vec{v}_{g}\right)\) + (M-dm)\((\vec{v}+d \vec{v})\) …….(2)

Here, the ejected gas and rocket constitutes a system at time (t + dt).
External force on the rocket is weight (-mg) of the rocket (the upward direction is taken as positive)
Now Impulse = Change is momentum

term dm d\(\vec{v}\) can be dropped as this product is negligîble in comparison of other two terms.
Thus we have

In equation (iii) \(\frac{d \vec{V}}{d t}\) represent the acceleration of the rocket, so M\(\frac{d v}{d t}\) = resultant force on the rocket.
Therefore,
Resultant force on rocket = Upthrust on the rocket – Weight of the rocket
Here the upthrust on the rocket is proportional to both the relative velocity \(\left(\overrightarrow{v_{g}}\right)\) of the ejected gas and the mass of the gas ejected per unit time \(\left(\frac{d m}{d t}\right)\).
Again from eq. (3)
\(\frac{d \vec{v}}{d t}=\frac{\overrightarrow{v_{g}}}{m} \frac{d m}{d t}\) – g …..(4)
As the rocket goes higher and higher, value of the acceleration due to gravity ‘g’ decreases continuously.

The values of (V_g) and (dm/dt). Practically remains constant while fuel is being consumed but remaining mass decreases continuously. This result increases in acceleration continues until all the fuel is burnt up.

The velocity of the rocket at any time ‘t’:
Now we will find out the relation between the velocity at any time ‘t and remaining mass. Again from equation (4) we have
\(d \vec{v}=\overrightarrow{v_{g}}\left(\frac{d m}{M}\right)\) – gdt ……….(5)
Initially at time t O if the mass and velocity of the rocket are m₀ and u₀ respectively. After time ‘t’, if m and u are the mass and velocity of the rocket. On integrating equation (5) within these limits

Note: Here dm is a quantity of mass ejected in time ‘dt’ so the change in mass of the rocket in time dt is – dm that’s why we have changed the sign of dm in equation (6).

On evaluating this integral we get

Equation (7) gives the change in velocity of the rocket in terms of exhaust speed and ratio of initial and final masses at any time ‘t’.

At time t = 0 the velocity of the rocket (initial velocity)

Note: The speed acquired by the rocket when the whole of the fuel is burnt out is called the burn-out speed of the rocket.

Physics Notes

Second Law of Motion is the real law of Motion Physics Notes

Second Law of Motion is the real law of Motion:
(A) First law is contained in the second law:
According to Newton’s second law of motion, the force acting on a body is given by:

Thus there is no force is applied on the body then the body at rest will remains at rest and a body in uniform motion will continue to move uniformly along the same straight path. Hence first law of motion is contained in the second law.

(B) Third law is contained in the second law: consider an isolated system of two bodies A and B suppose the two bodies interact mutually with one another, Let \(\vec{F}_{B A}\) be the force (action) exerted by A or B.

Let \(\frac{d \overrightarrow{p_{B}}}{d t}\) be the resulting charge of momentum of B.

Let \(\vec{F}_{B A}\) be the force (reaction) exerted by B or A Let \(\frac{d \overrightarrow{p_{A}}}{d t}\) be the resulting change of momentom of A

According to Newtons IInd law of motion:

in the absence of external force, the rate of change in momentum must be zero.
\(\overrightarrow{F_{B A}}+\overrightarrow{F_{A B}}\) =0 ⇒ \(\vec{F}_{B A}=-\vec{F}_{A B}\)
This indicate Newton’s IIIrd law of motion we can say that Newton’s IIIrd law is the real law of motion.

Physics Notes

Principle of Conservation of Linear Momentum and its Applications Physics Notes

Principle of Conservation of Linear Momentum and its Applications:
If the net external force acting on a system of bodies is zero, then the momentum of the system remains constant. This is the basic principle of conservation of linear momentum.

According to Newton’s second law

Considering external force on the particle (or a body)= zero
we have \(\vec{F}=\frac{d \vec{p}}{d t}\) = 0
⇒ \(\vec{p}\) = constant …(1)
If net force (or the vector sum of all forces) on system of particle is equal to zero, the vector sum of linear momentum of all particles remains conserved.

Consider a system of two bodies on which no external force acts. The bodies can mutually interact with each other. Due to the mutual interaction of the bodies, the momentum of the individual bodies may increase or decrease according to the situation, but the momentum of the system will always be conserved, as long as there is no external net force acting on it.

Thus, if \(\vec{p}_{1}\) and \(\vec{p}_{2}\) are momentum of the two bodies at any instant, then in absence of external force
\(\vec{p}_{1}+\vec{p_{2}}\) = constant …….(3)

If due to mutual interaction, the momentum of two bodies becomes p{ and p2 respectively, then according to the principle of conservation of momentum

Where \(\vec{u}_{1}\) and \(\vec{u}_{1}\) are initial velocities of the two bodies of masses m₁ and m₂ and v₁ and v₂ are their final velocities.

Therefore, the principle of conservation of linear momentum may also be stated as follows:
For an isolated system (a system on which no external force acts), the initial momentum of the syster A is equal to the final momentum of the system.

Practical Applications of the Principle of Conservation of Momentum:
1. Recoiling a Gun: Let’s consider the gun and bullet in its barrel as an isolated system. In the beginning when the bullet is not fired both the gun and the bullet are at rest. So the momentum before firing is zero
or \(\overrightarrow{p_{c}}\) = 0
Now when the bullet is fired, it moves, in the forward direction and gun recoils back in the opposite direction.

Let m_b be the mass and v_b be the velocity of the bullet and m_g and v_g be the mass and velocity of the gun after firing.

Total momentum of the system after the firing would be
\(\overrightarrow{p_{f}}=m_{b} \overrightarrow{v_{b}}+m_{g} \overrightarrow{v_{g}}\)
Since, no external forces are acting on the system, we can apply the law of conservation of linear momentum therefore,

Total momentum of gun and bullet before firing = Total momentum of gun and bullet after firing
0 = m_b \(\overrightarrow{v_{b}}\) + m_g \(\overrightarrow{v_{g}}\)
or
\(\overrightarrow{v_{g}}=-\frac{m_{b} \overrightarrow{v_{b}}}{m_{g}}\)

The negative sign shows that v_g and v_b are in opposite directions i.e., as the bullet moves forward, then the gun will move in the backward direction. The backward motion of the gun is called the recoil of the gun.

2. While firing a bullet, the gun must be held tight to the shoulder: This would save hurting the shoulder of the man who fires the gun as the recoil velocity of the gun. If the gun is held tight to the shoulder, then the gun and the body of the man recoil as one system. As the total mass is quite large, the recoil velocity will be very small and the shoulder of the man will not get hurt.

3. Rockets work is on the principle of conservation of momentum: The rocket’s fuel burns and pushes the exhaust gases downwards, due to this the rocket gets pushed upwards. Motorboats also work on the same principle, it pushes the water backwards and gets pushed forward in reaction to conserve momentum.

Physics Notes

Newton’s Third Law of Motion Physics Notes

Newton’s Third Law of Motion:
If an object ‘A exerts a force on object ‘B, then object B must exert a force of equal magnitude and opposite direction back on object A.

This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as action-reaction, where the force exerted is the action and the force experienced as a consequence is the reaction.

According to Newton’s third law, “To every action, there is always an equal and opposite reaction”.

It must be remembered that action and reaction always act on different objects. The third law of motion indicates that when one object exerts a force on another object, the second object instantaneously exerts a force back on the object. These two forces are always equal in magnitude, but opposite in direction.

These forces act on different objects and so they do not cancel each other. Thus Newton’s third law of motion describes the relationship between the forces of interaction between two objects.

For example, when we placed a wooden block on the ground, this block exerts a force equal to its weight, W = mg acting downwards to the ground. This is the action force. The ground exerts an equal and opposite force N = mg on the block in upward direction. This is the reaction force.

Action weight of the body acting downwards

Illustrations of Newton’s Third Law:
Some of the examples of Newton’s third law of motion are given below:
1. A gun recoils when a bullet is fired form it: When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun it moves only a

little backward by giving a jerk at the shoulder of the gun man. The backward movement of the gun is called the recoil of the gun.

2. Walking: In order to walk, we press the ground in backward direction with our feet (action). In turns, the ground gives an equal and opposite reaction R, (figure (a)). The reaction R can be resolved into two components, one along the horizontal and other along the vertical. The component H = R cosθ along the horizontal, help us to move forward, while the vertical component, V = R sinθ opposes our weight, [figure (b)]

3. Swimming: While swimming, the swimmer pushes the water backward (action). The water pushes the swimmer forward (reaction) with the same force.

The swimmer pushes down and backwards against the water.

4. Flight of jet planes and rockets: Rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas, therefore, exerts a large reaction force forward on the rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases.

5. The flying of a bird: The wings of a bird force air a downward direction. In turn, the air gives equal and opposite reactions to the wings. The resultant reaction acts on the bird in an upward direction and makes the bird to fly upward.

Similarly, the Helicopter creates lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need.

6. It is difficult to walk on sand or ice: As we press the sandy ground in the backward direction, the sarid gets pushed away and as a result, we get only a very small reaction from the sandy ground, making it difficult for us to walk. In case of ice, due to little friction between our feet and ice surface, there is hardly any forward reaction and hence we cannot walk on it.

7. Rebounding of a rubber ball: When a rubber is struck against a wall or floor, it exerts a force on wall (action). The ball rebounds with an equal force (reaction) exerted by the wall or floor on the ball.

Important Facts:

1. Forces of action and reaction act always on different bodies. Hence, they never cancel each other. Each force produces its own effect.
If we consider a pair of bodies A and B. then according to third law,
\(\vec{F}_{A B}=-\vec{F}_{B A}\)
i.e force on A by B= – force on B by A

2. If action and reaction forces were to act on the same body, their resultant would be zero.

3. It is wrong impression that action comes before reaction i. e., the action is the cause and reaction is the effect. The fact is that the two act at the same instant.

4. Newton’s third law is applicable for both kind of bodies i.e.. for bodies at rest or they are in motion.

5. Forces always occur in pairs.

6. The third law applies to all types of forces, e.g, gravitational, electric or magnetic forces etc.

Physics Notes

Impulse and Impulse-Momentum Theorem Physics Notes

Impulse and Impulse-Momentum Theorem:
A force which acts on a body for short interval of time is called impulsive force or impulse.

For Example Hitting, jumping, diving, catching etc. are all examples of impulsive forces or impulse.

An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. Thus it is not possible to measure easily the value of impulsive force because it changes with time. In such cases, we measure the total effect of the force, called impulse hence, impulse is defined as the product of the average force and the time interval for which the force acts.

If \(\vec{F}\) is the value of force during impact at any time and \(\vec{p}\) is the momentum of the body at that time, then according to Newton’s second law of motion,

\(\vec{F}\) = \(\frac{d \vec{p}}{d t}\)
or \(\vec{F}\) dt = d\(\vec{p}\)

Suppose that the impact lasts for a small time t and during this time, the momentum of the body changes from \(\vec{p}_{1}\) to \(\vec{p}_{2}\) then integrating the above equation between the proper limits, we have:

It may be noted that F has not been taken out of the integration sign for the reason that varies with time and does not remain constant during impact. The integral ∫t0 F dt is measure of the impulse, when the force of impact acts on the body and from equation (1) we find that it is equal to total change in momentum of the body. Since impulse is equal to a scalar (time) times a vector (force) or equal to the change in momentum (vector), it is a vector quantity and it is denoted by I.

However if \(\vec{F}_{a v}\) is the average force (constant) during the impact, then

i. e., the change in momentum of an object equals the impulse applied to it. This statement is called
impulse-momentum theorem.
I = Δp
Dimensional Formula and Unit:
I = FΔt = [M¹L¹T^-2] |T¹]= [M¹L¹T^-1]
So, the dimensional formula of impulse is the same as that of momentum.

The SI unit of impulse are (N-s) and kg m/s.
In C.G.S. system unit of impulse are dyne-sec and g cm/s.

Force Time Curve: In the real world, forces are often not constant. Forces may build up from zero over time and also may vary depending on many factors.

Finding out the overall effect of all these forces directly would be quite difficult. As we calculate impulse, we multiply force by time. This is equivalent to finding out the area under a force time curve. For variable force the shape of the force-time curve would be complicated but for a constant force we will get a simpler rectangle. In any case, the overall net impulse only matters to understand the motion of an object following an impulse.

In the figure, the graph of change in impulsive force with the time is shown:

The force-time curve and the area between the time axis can be divided in the form of many slabs. Suppose the value of force F is considered as constant along the change in time dt, then area of slab is given by F. dt.

The total effect of the force for time t₁ to t₂
= ∫^t₀ Fdt = sum of area of all slabs o
= graph of force-time and area covered between the time axis
∵ By the impulse-momentum theorem t
Impulse I = ∫^t₀ Fdt = p₂ – p₁ = change in momentum.

Thus force-time graph and the area covered with the time axis is equal to the total change in the momentum of the body.

Applications of the Concept of Impulse:
1. While catching a ball, a cricket player lowers his hands to save himself from getting hurt:

By lowering his hands, the cricket player increases the time interval in which the catch is completed. As the total change in momentum takes place in a large time interval, the time rate of change of momentum of the ball decreases. So, according to Newton’s second law of motion, lesser force acts on the hands of player and he saves himself from getting hurt.

2. Cars, buses, trucks, bogies of train etc. are provided with a spring system (shockers) to avoid severe jerks:
When they move over an uneven road, impulsive forces are exerted by the road. The function of shockers is to increase the time of impact. This would reduce the force/jerks experienced by the rider of the vehicle.

3. An athlete is advised to come to stop slowly: After finishing a fast race an athlete is advised to come to stop slowly, so that time of stop increases and hence force experienced by him decreases.

4. In a head-on collision: Between two vehicles change in linear momentum is equal to sum of the linear momentum of the two vehicles. As the time of impact is small, an extremely large force develops which causes damage to the vehicle.

Physics Notes

Momentum and Newton’s Second Law of Motion Physics Notes

Momentum and Newton’s Second Law of Motion:
Linear momentum: Momentum of the body is the physical quantity of motion possessed by the body and mathematically, It is defined as the product of mass and velocity of the body.

As the linear momentum or simply momentum is equal to a scalar time a vector (velocity), it is
therefore a vector quantity and is denoted by p. The momentum of a body of mass m moving with velocity v is given by the relation:
\(\vec{p}=m \vec{v}\)
Dimensional formula = [M¹L¹T^-1]
Unit = Kg m/s

Suppose that a ball of mass mi and a car of mass m₂ (m₂ > m₁) are moving with the same velocity v. If p₁ and p₂ are momentum of ball and car respectively then:
\(\frac{p_{1}}{p_{2}}=\frac{m_{1} v}{m_{2} v}\)
or
\(\frac{p_{1}}{p_{2}}=\frac{m_{1}}{m_{2}}\)

As m₂ > m₁: It follows that p₂ > p₁. If a ball and a car are traveling with the same velocity, the momentum of the car will be greater than that of the ball. Similarly, we can show that if two objects of same masses are thrown at different velocities, the one moving with the greater velocity possesses greater momentum. Finally, if two objects of masses m₁ and m₂ are moving with velocities v₁ and v₂ possess equal momentum.
m₁v₁ = m₂ v₂
\(\frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}}\)
In case m₂ > m₁ then v₂ < v₁ i. e, two bodies of different masses possess same momentum, the lighter body possesses greater velocity.

The concept of momentum was introduced by Newton in order to measure the quantitative effect of force.
The momentum of body in terms of kinetic energy

Explanation of Newton’s Second Law:
According to Newton’s second law of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force on the body, and this change takes place always in the direction of the applied force.
Let, m = mass of a body
v = velocity of the body
The linear momentum of the body
\(\vec{p}=m \vec{v}\) …(1)
let \(\vec{F}\) = External force applied on the body in the direction of motion of the body. .
Δ \(\vec{p}\) = a small change in linear momentum of the body in a small time Δt.
Rate of change of linear momentum of the body = \(\frac{\Delta \vec{p}}{\Delta t}\)

According to Newton’s second law

Where k is proportionality constant.
Taking the limit Δt → 0, the term \(\frac{\Delta \vec{p}}{\Delta t}\) becomes the derivative or differential coefficient of \(\vec{p}\) w.r.t. time t.
It is denoted by \(\frac{d \vec{p}}{d t}\)
\(\vec{F}=k \frac{d \vec{p}}{d t}\)
Where k = 1 in all the system

As acceleration is a vector quantity and mass is scalar, therefore force \(\vec{F}\) being the product of m and \(\vec{a}\) is a vector. The direction of is the same as the direction of \(\vec{a}\). Equation (4) represents the equation of motion of the body. We can rewrite equation (6) in scalar form as:
F = ma …(7)
Thus, magnitude of the force can be calculated by multiplying mass of the body and the acceleration produced in it. Hence, second law of motion gives a measure of force.

Important Facts:
1. If the applied force produces acceleration a, such that ax,ay and a2 are the magnitudes of the component of acceleration along the X-axis, Y-axis, X-axis respectively.
Then
\(\vec{F}\) = m(a_x î + a_y ĵ + a_z k̂) …(iv)
If F_x, F_y ,F_z are components of force along X-axis, Y-axis and X-axis respectively, then
\(\vec{F}\) = F_xî + F_y ĵ + F_z k̂ …….(v)

From the equation (4), (5) we have

The set of equation (8) expresses Newton’s second law of motion in component form. Three mutually perpendicular components of the force and the acceleration have to obey the set of equation (8).

As the force is equal to a scalar (mass) times a vector (acceleration), it is a vector quantity. The equation (8) is called the equation of motion of body.

In scalar form F = ma
If we know the values of m and a, the force F acting on the body can be calculated and hence second law gives a measure of the force.

2. Equal forces applied for equal time on different bodies change equal momentum.

3. Newton’s second law shows the relation between net external force and acceleration of the body.

4. Initially two bodies are in rest position, a constant force is applied for a definite time interval then the lighter body receives more speed than the heavy body because of change in momentum In both bodies are equal.
Thus, m₁v₁ = m₂v₂
If m₁ < m₂
then v₁ > v₂

Dimension and Unit of Force:
As F = ma
.-. F= [M¹] [L¹T^-2 ] = [M¹L¹T^-2]
This is the dimensional formula of force.

Unit of force:
The unit of force is Newton in the (M.K.S.) system and dynein the C.G.S. system and Poundal in (F.P.S.) system.
Definition of 1 Newton
F = ma
When m = 1kg and a = 1 m/s
then F = 1 Newton
The force which produces the acceleration of 1m/s^-2 in the 1 kg body, is equal to the 1 Newton.
1 Newton = 1 kg m/ s²

In C.G.S. System
F = ma
If m = 1 g and a = 1 cm/s²
then F = 1 dyne
Thus 1 dyne is that force that produces 1 cm/s acceleration in the mass of 1g body.

Physics Notes