Circular Motion

Uniform circular motion

Let us consider an object of mass* m* tied to an inextensible string and whirled in a vertical circle of radius *r*. Then, due to the earth's gravitational field, its velocity varies from its maximum value *v _{2}_{ }*at the lowest point B to its lowest value

*v*at the highest point A

_{1 }*.*

Let us make the following assumptions:

A = Highest position

B = Lowest position

C = Midway position

*v*= Velocity at highest point A

_{1}*v*= Velocity at lowest point B

_{2}*v*= Velocity at midway point C

_{3}*T*

_{1}

_{ = }Tension when object is at A

*T*

_{2}

_{ = }Tension when object is at

*B*

The various forces acting on the body when it is at the highest position A are:

(1) Weight

*mg*acting vertically downward

(2) Tension

*T*

_{1}acting vertically downward

Net downward force acting on the body =

*mg + T*

_{1}This force will provide the necessary centripetal force for the body to move in the circle.

${T}_{1}+mg=\frac{m{{v}_{1}}^{2}}{r}\phantom{\rule{0ex}{0ex}}\therefore {T}_{1}=\frac{m{{v}_{1}}^{2}}{r}-mg\phantom{\rule{0ex}{0ex}}$

The body will perform circular motion only when

*T*is greater than or equal to zero. If

_{1}*T*

_{1}is less than zero the string will slacken and the body will fall down. The velocity of the body at the highest point can be calculated as:

When (

*T*

_{1}=0)

$\therefore \frac{m{{v}_{1}}^{2}}{r}=mg\phantom{\rule{0ex}{0ex}}\therefore {v}_{1}=\sqrt{rg}.....\left(1\right)$

Total energy of the particle at A = K.E at A + P.E at A

$\Rightarrow E=\frac{1}{2}m{{v}_{1}}^{2}+2mgr$

Using equation (1) we get:

$E=\frac{1}{2}mgr+2mgr\phantom{\rule{0ex}{0ex}}\therefore E=\frac{5}{2}mgr.....\left(2\right)\phantom{\rule{0ex}{0ex}}$

Similarly, various forces acting on the body at B will be

(1) Tension

*T*

_{2}acting vertically upward

(2) Weight

*mg*acting vertically downward

Net upward force acting on the body =

*T*

_{2}$-$ mg

This will provide the required centripetal force to the body.

${T}_{2}-mg=\frac{m{{v}_{2}}^{2}}{r}\phantom{\rule{0ex}{0ex}}\therefore {T}_{2}=\frac{m{{v}_{2}}^{2}}{r}+mg$

This is the tension acâ€¦

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