A body of mass 'm' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is
( $g$ = gravitational acceleration)

Question

TestHub · Accepted Answer

For a body to just complete a vertical circle, the tension in the string at the highest point is zero.

$T_{H} + m g = \frac{m v _{H}^{2}}{r} \Rightarrow v_{H} = g r$

By conservation of mechanical energy between the highest and lowest points:

$\frac{1}{2} m v_{L}^{2} = \frac{1}{2} m v_{H}^{2} + m g (2 r)$

$v_{L}^{2} = v_{H}^{2} + 4 g r = g r + 4 g r = 5 g r$

At the lowest point, the net force towards the center provides the centripetal acceleration:

$T_{L} - m g = \frac{m v _{L}^{2}}{r}$

$T_{L} = m g + \frac{m ( 5 g r )}{r} = m g + 5 m g = 6 m g$

The apparent weight of the body is equal to the tension in the string at that point, which is $6 m g$ .

Answer: $6 m g$

Physics - W.E.P. Question with Solution | TestHub

Options:

Doubts & Discussion