Circular Motion in Horizontal and Vertical Planes Physics Notes

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