A disc is rolling (without slipping) on a horizontal surface C is its centre and Q and P are two points equidistant from C. Let vP, vQ and vC be the magnitude of velocities of points P, Q and C respectively, then
During pure rolling the point on ground acts as instantaneous axis of rotation. Distance from the point of contact with ground determines speed of point. so,
\[ V_{Q}> V_{C}> V_{P} \]
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is
\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \]
\[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \]
Solving I = MR²/2 so, Object is a disc or hollow cylinder.
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.