Prove the result that the velocity v of ...

Systems of Particles and Rotational Motion

Question : 27 of 33

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Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by

v^{2}= \frac{2 g h}{1+\frac{k^{2}}{R^{2}}}

using dynamical consideration (i.e. by consideration of forces and torques).

Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Solution:

Let M, R, k be the mass, radius and radius of gyration of a body placed at the top A of the inclined plane of height h and angle of inclination θ.

∴ Its total energy at point

A = \text{P.E.} = mgh

Its total energy at point

B = K.E. of translation + K.E. of rotation

= \frac{1}{2} m v^{2}+ \frac{1}{2} I \omega^{2}

According to the principle of conservation of energy we have

\frac{1}{2} m v^{2}+ \frac{1}{2} (m k^{2}) \frac{v^{2}}{R^{2}} = m g h \left[\because I=m k^{2}, \omega = \frac{v}{R}\right]

\Rightarrow \frac{1}{2} m v^{2} \left(1+ \frac{k^{2}}{R^{2}}\right) = m g h

\therefore v^{2}= \frac{2 g h}{1+\frac{k^{2}}{R^{2}}}

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