Question

A spherical ball of mass $m$ and radius $r$ rolls without slipping on a rough concave surface of large radius . It makes small oscillations about the lowest point. Find the time period.

Answer 1

It is given that a small ball of radius $r$rolls around the surface of a rough concave surface of $R$ radius at its lowest points. Assume the distance between the ball and the center of the concave surface to be $R - r$. Using equations of simple harmonic motions to solve for the angular velocity and using that to find the time period of oscillation.

Complete step by step solution:
Let us assume that the ball undergoes some angular velocity at the lowest points of the concave rough surface. This angular velocity is represented as $\omega$. Now, the centripetal velocity of the system is given using the formula,
$\Rightarrow v = r\omega$
We can say that the distance traversed by the ball is different between its radii’s. Substituting this on the above equation we get,
$\Rightarrow v = (R - r)\omega$
Let ${\omega _0}$be the angular velocity of the spherical ball. Now, it’s velocity is equal to that of the velocity of the system, since the system is centripetal.
$\Rightarrow v = (r){\omega _0}$
$\Rightarrow {\omega _0} = \dfrac{{R - r}}{r}\omega$
Total energy in a simple harmonic motion is said to be constant. Using this we can say the energy experienced by the ball and the system sums up to a constant value. Assuming the ball is away from the center of the concave surface in angle $\theta$, we get the sum of potential energy of the system , the kinetic energy of the system and the kinetic energy of the ball as constant.
$\Rightarrow mg(R - r)(1 - \cos \theta ) + \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega _0}^2 = Con$
On substituting for angular velocity and${\omega _0}$, we get,
$\Rightarrow mg(R - r)(1 - \cos \theta ) + \dfrac{1}{2}m{(R - r)^2}{\omega ^2} + \dfrac{1}{2} \times \dfrac{2}{5}m{r^2}{(\dfrac{{R - r}}{r})^2}{\omega ^2} = Con$
Cancelling out the common terms , we get,
$\Rightarrow mg(R - r)(1 - \cos \theta ) + \dfrac{1}{2}m{(R - r)^2}{\omega ^2} + \dfrac{1}{5}m{(R - r)^2}\omega$
Since RHS is constant, we can take out mass and assume it as constant.
$\Rightarrow g(R - r)(1 - \cos \theta ) + \dfrac{1}{2}{(R - r)^2}{\omega ^2} + \dfrac{1}{5}{(R - r)^2}{\omega ^2} = Con$
$\Rightarrow g(R - r)(1 - \cos \theta ) + (\dfrac{7}{{10}}){(R - r)^2}{\omega ^2} = const$
Differentiating with respect to time t , we get,
$\Rightarrow g(R - r)( - \sin \theta )\dfrac{{d\theta }}{{dt}} + (\dfrac{7}{{10}}){(R - r)^2}2\omega \dfrac{{d\omega }}{{dt}} = const$
Taking the negative term to the other side we get,
Cancelling out the common term, we get,
$\Rightarrow g(R - r)( - \sin \theta )\dfrac{{d\theta }}{{dt}} = (\dfrac{7}{{10}}){(R - r)^2}2\omega \dfrac{{d\omega }}{{dt}}$( Since $\dfrac{{d\omega }}{{dt}}$is equal to angular acceleration )
$\Rightarrow g(\sin \theta )\omega = (\dfrac{7}{{10}})(R - r)2\omega \alpha$(Since is equal to angular velocity)
Cancelling out the common term , we get,
$\Rightarrow g(\sin \theta ) = (\dfrac{7}{5})(R - r)\alpha$
On, rearranging we get,
$\Rightarrow \dfrac{{5g(\sin \theta )}}{{7(R - r)}} = \alpha$
Rounding off $\sin \theta$to $\theta$, we get,
$\Rightarrow \dfrac{{5g(\theta )}}{{7(R - r)}} = \alpha$
$\Rightarrow \dfrac{{5g}}{{7(R - r)}} = \dfrac{\alpha }{\theta }$
We know that angular acceleration and the displacement ratio to be square of angular velocity, thus
$\Rightarrow \dfrac{{5g}}{{7(R - r)}} = {\omega ^2}$
$\Rightarrow \omega = \sqrt {\dfrac{{5g}}{{7(R - r)}}}$
We know , time period of oscillation is inverse of its angular velocity, thus
$\Rightarrow T = 2\pi \sqrt {\dfrac{{7(R - r)}}{{5g}}}$

Note: In the energy equation , we consider the moment of inertia of a sphere for kinetic energy of the spherical ball because it is assumed that the ball is undergoing centripetal acceleration and is rotating, thus can’t have an equation that considers normal velocity v.