Question

A thick-walled hollow sphere has outer radius R. It rolls down an inclined plane without slipping and its speed at the bottom is v. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom will be 5v/4. What is radius of gyration of the sphere?

Answer 1

Hint:For pure rolling, the potential energy gets converted into rotational kinetic energy and translational kinetic energy. For pure slipping, the potential energy of the solid at the top of the inclined plane gets converted only into translational kinetic energy as there will be no rotational motion in that case.

Formula Used:
Potential Energy, \[P.E. = mgh\]
where m = mass of the sphere and g = acceleration due to gravity
Total kinetic energy,
\[K.E. = K.E{._{rot}} + K.E{._{tran{s_{}}}}\]
\[\Rightarrow K.E. = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]
For no slipping, the condition is:
\[v = \omega R\]
where v = linear velocity, \[\omega \]= angular velocity, I = moment of inertia and R is the radius of the sphere.

Complete step by step solution:
Given: A hollow sphere (let its mass be m) of outer radius R with its velocity \[v\]rolls down the inclined plane without slipping. When the inclined plane is frictionless, the new velocity is 5v/4.

Case-1: For case when velocity = v, Let the angular velocity be \[\omega \]. Let the coefficient of friction be \[\mu \]. For no slipping, the condition is:
\[v = \omega R\] ---- (1)
Let the height of the inclined plane from the ground be h and moment of inertia of the sphere about its own axis be I. Using Conservation of energy as there is no slipping, the loss in potential energy is equal to translational and rotational Kinetic energy. Then,
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\]---- (2)
Using values from equation (1), we get,
\[mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}{\dfrac{{Iv}}{{{R^2}}}^2}\]---- (3)

Case-2: For case when velocity = 5v/R,
If there is no rotation then the potential energy is converted to translational kinetic energy. Applying conservation of energy,
\[mgh = \dfrac{1}{2}m{\left( {\dfrac{{5v}}{R}} \right)^2}\]---- (4)
Equating (3) and (4)
\[\dfrac{1}{2}m{v^2} + \dfrac{1}{2}{\dfrac{{Iv}}{{{R^2}}}^2} = \dfrac{1}{2}m{\left( {\dfrac{{5v}}{R}} \right)^2}\]
\[\Rightarrow I = \dfrac{9}{{16}}m{R^2}\]---- (5)
\[\Rightarrow I = m{k^2}\]----(6)
where k is the radius of gyration.
Comparing equations (5) and (6), we get,
\[\therefore k = \dfrac{3}{4}R\]

Hence the radius of gyration is \[\dfrac{3}{4}R\].

Note: For pure rolling, the rotational acceleration and the acceleration of the centre of mass should have an equivalence of the form \[R\alpha = {a_{CM}}\]. For pure slipping, there will be no rotational motion. Hence the angular velocity and hence the rotational kinetic energy becomes zero. In the real world, friction is important so that there is no slipping as frictional force will act in opposite direction to the translational motion and thus prevent slipping.