Rolling motion

Question

A body initially at rest and sliding along a frictionless track from a height *h* (as shown in the figure) just completes a vertical circle of diameter *AB=D*. The height *h* is equal to

Moderate

Solution

As body is at rest initially, i.e., speed=0

At point *A*, speed*=v* As track is frictionless, so total mechanical energy will remain constant.

$\therefore (\mathrm{T}.\mathrm{M}.\mathrm{E}{)}_{i}=(\mathrm{T}.\mathrm{M}.\mathrm{E}{)}_{f}$

$0+mgh=\frac{1}{2}m{v}^{2}+0\text{or}h=\frac{{v}^{2}}{2g}$

$\text{For completing the vertical circle,}v\ge \sqrt{5gR}$

$\therefore h=\frac{5gR}{2g}=\frac{5}{2}R=\frac{5}{4}D$

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