Discussion Forum

Two solid cylinders P and Q of the same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Answer:

Explanation:

$I_{P} >I_{Q}$

 In case of pure rolling,

   $a=\frac{g \sin \theta}{1+I/mR^{2}}$

$a_{Q} >a_{p}$  as its moment of inertia is less.

Therefore, Q reaches first with more linear speed and more translational kinetic energy.

 Further, $\omega=\frac{v}{R}$

 or   $\omega \propto V$

$\therefore$   $\omega_{Q} >\omega_{P}$                as   $v_{p} >v_{Q}$