First slide

Vertical circular motion

Question

A particle of mass m slides from rest from position A, along a smooth circular track of radius R as shown in fig. (7a). The vertical distance travelled by m before leaving the surface of the track is

Moderate

A

R/3
B

R/2
C

3/R
D

2/R

Solution

The speed y acquired by the particle after falling through a distance h (from A to B) is given by
$\frac{1}{2} {mv}^{2} = mgh$
The contributed force needed at this instant is
$\begin{array}{l} \frac{{mv}^{2}}{R} = \frac{m â 2 gh}{R} \\ or v = \sqrt{2 gh} \\ Now mgcos â¡ Î¸ = \frac{{mv}^{2}}{\overset{Â¯}{R}} \\ or R â h = 2 h \\ â´ h = R / 3 \end{array}$

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