Question

1. A body of mass m attached to a thread is revolved along a
vertical circle of radius r. If its vclocity at the topmost point of
the circle is v, the tension of the thread at the instant will be:
(a)mg-mv^2r
(b)mv^2/r
(c)(mv^2/r)-mg
(d)(mv^2/r)+mg​

Answer 1

Given:

• The body of mass m is revolving in a vertical circle of radius r.

• Velocity at topmost point of the circle is v.

To find:

• Tension at that instànt?

Calculation:

We know that, when the object is moving in a vertical circle the net force experienced by the object will always be equal to the centripetal force.

$\therefore \: F_{net} = F_{c}$

• Now, the force can be written in terms of weight of the ball and the tension of the string.

$\implies \: W + T= F_{c}$

$\implies \: mg + T= \dfrac{m {v}^{2} }{r}$

$\boxed{ \implies \: T= \dfrac{m {v}^{2} }{r} - mg}$

So, net tension of string is :

$\boxed{ \bf \: T= \dfrac{m {v}^{2} }{r} - mg}$

• Option C) is correct ✔️

Answer 2

C) $(\frac{mv^2}{r})-mg$ is the tension.

Step-by-step Explanation:

Given: mass of the body = m

The radius of the verticle circle = r

The velocity of the topmost point = v

To Find: Tension of the thread

Solution:

• Finding tension of the thread

Consider a body on mass 'm' attached to a thread is revolved along with a vertical circle of radius 'r' such that the centripetal force is,

$F_c = \frac{mv^2}{r}$

At the topmost position, the centripetal force 'Fc' is balanced by the tension 'T' and gravitation force 'mg' such that

$\frac{mv^2}{r} = T + mg$

Therefore, the tension of the thread is,

$T = \frac{mv^2}{r} - mg$

Hence, $T = (\frac{mv^2}{r})-mg$ is the tension of the thread.