Motion of a body in a Vertical and Horizontal Circle

Subject: Physics

Overview

A body of mass m is tied with a string of length l and then revolve around in the verticle circle of radius (r) is equal to the length of the string . If a body is tied by a weightless and inextensible string over the rigid support and moves in a horizontal circle then the body is called conical pendulum.

Motion of a body in a Vertical Circle

Consider a body of mass m is tied with string of length l and then revolve around in the verticle circle of radius (r) is equal to the length of the string,

At any instant of time the body is at P such that string makes $\angle \theta$ with vertical. The weight (mg) of a body is acted downward which can be resolved into two components. Horizontal component $mg \cos \theta$ is balanced by the tension where remaining force provides the necessary centripetal force by which object is moved in a vertical centre.

$$\text {i.e.} T-mg\cos\theta = \frac {mv^2}{r}$$

$$\text {or,} T = \frac {mv^2}{r} +mg\cos\theta \dots (i)$$

Special cases

1. For a body at A (lowest position)
$$\theta = 0^o$$
$$\text {or,} T = \frac {mv_1^2}{r} +mg\cos 0^o$$
$$\therefore \:T_{max} = \frac {mv_1^2}{r} +mg$$
$$= m\omega ^2 r + mg \dots (ii)$$

2. For a body at B
$$\theta = 90^o$$
$$\text {or,} T = \frac {mv_1^2}{r} +mg\cos90^o$$
$$\therefore \:T_{max} = \frac {mv_1^2}{r} = m \omega^2 r \dots (iii)$$

3. For a body at C (highest position)
$$\theta = 180^o$$
$$\text {or,} T = \frac {mv_1^2}{r} +mg\cos180^o$$
$$\therefore T = \frac {mv_1^2}{r} - mg \dots (iv)$$

4. For a body at C (highest position)
$$\theta = 360^o$$
$$\text {or,} T = \frac {mv_1^2}{r} +mg\cos360^o$$
$$\therefore T = \frac {mv_1^2}{r} \dots (v)$$

At highest point C

Let V1 is the minimum velocity

$$\text {or,} T = \frac {mv_1^2}{r} +mg\cos 0^o$$

$$\text {If}\:T = \frac {mv_1^2}{r} +mg \: , T>0$$

Body is continuing in a vertical circle

$$\text {If}\: mg > \frac {mv_1^2}{r} \: , T>0$$

Body fall down

for just remain at points, T = 0

$$\text {or,} \frac {mv_1^2}{r} -mg = 0$$

$$\text {or,} \frac {mv_1^2}{r} = mg$$

$$v_1 = \sqrt {rg}$$

For point A

Let V2 is the main velocity of a body in a vertical circle at A. Using a body in a vertical circle at A.

Using conservation of energy

$$\text {Total energy at A} = \text {Total energy at C}$$

$$\text {Kinetic energy at A} = \text {Potential energy} + \text {Kinetic energy at C}$$

$$\frac 12 m v_2^2 = \frac 12 m v_1^2 + mg (2r)$$

$$\frac {v_2^2}{2} = \frac 12 v_1^2 + 2gr$$

$$\frac {v_2^2}{2} = \frac 52 gr$$

$$\boxed {V_2 = \sqrt {5gr}}$$

Motion of a body in a Horizontal Circle (conical pendulum)

If a body is tied by a weightless and inextensible string over the rigid support and moves in a horizontal circle then the body is called conical pendulum.

Let us consider a body of mass with the string with length (l) over the rigid support Q and whirl in the horizontal circle of radius (r) at any instant of time body is at P where weight (mg) of body tension (T) is acted along the string towards the centre which can be resolved into component vertical Tcosθ balance mg and horizontal Tsinθ provides necessary centripetal force.

$$\text {i.e.} T\cos\theta = mg \dots (i)$$

$$T\sin\theta = \frac {\cos V^2}{r} \dots (ii)$$

Dividing equation ii by i

$$\tan \theta = \frac {V^2}{rg}$$

$$\boxed {\theta = \tan ^{-1}\frac {V^2}{rg}}$$

Things to remember
• $$T-mg\cos\theta = \frac {mv^2}{r}$$
• $$T = \frac {mv^2}{r} +mg\cos\theta \dots (i)$$
• If a body is tied by a weightless and inextensible string over the rigid support and moves in a horizontal circle then the body is called conical pendulum.

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