# Rolling sphere placed on a horizontal surface

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### Problem

A solid sphere of mass $m$ and radius $r$ is spinning about a horizontal axis with an angular velocity $\omega_1$ is placed on a rough horizontal surface. Subsequently, it rolls without slipping with an angular velocity of $\omega_2$. Find $\frac{\omega_2}{\omega_1}$

### Attempt at solution

I applied the conservation of energy. Initially the energy due to rotational motion only, can be calculated by $$\frac{1}{2}I\omega_1^2$$ where $I$ is the moment of inertia of the sphere.

After placing the sphere on the table, the energy of the sphere is due to its rotational and transnational motion. This can be calculated by

$$\frac{1}{2}I\omega_2^2 + \frac{1}{2}mv_{cm}^2$$ where $v_{cm}$ is the velocity of the centre of mass of the sphere. Since the problem states the sphere rolls without slipping, $v_{cm} = \omega_2\cdot r$ On equating, I calculated $\frac{\omega_2}{\omega_1}$ as $\sqrt\frac{2}{7}$

### Issue

The answer given is $\frac{2}{7}$. I'm unable to find where I went wrong.

edited May 30, 2018

When the sphere touches the ground at first, its angular velocity $\omega$ and linear velocity $v$ are not related by $v=r\omega$. This means that there is slipping at first, so some kinetic energy is lost because of work done against friction. So kinetic energy is not conserved.

The question states that subsequently (ie after release) the sphere rolls without slipping. However, the linear velocity of the sphere cannot change instantaneously from zero to the final value $r\omega_2$ because that implies infinite acceleration, which is not possible.

See http://farside.ph.utexas.edu/teaching/301/lectures/node115.html. You must assume there is some finite value of friction. This value does not affect the final angular velocity but does affect the time it takes until the slipping stops.

This method gives the book value of $\frac{\omega_2}{\omega_1}=\frac27\approx 0.2857$ which is smaller than the prediction of $\sqrt{\frac27}\approx 0.5345$ obtained from conservation of energy. This is as it should be, because we know that some energy is lost due to slipping.

answered Oct 31, 2017 by (28,876 points)
edited May 30, 2018
@sammy gerbil Alright, so I understand that my application of conservation of energy is incorrect as I haven't taken into account the energy lost due to friction. But since the question does not provide the value of frictional coefficient, how would I approach the solution for this problem? I have angular momentum in mind, but not sure how I would apply that.
Follow the method used in the link in my answer. The coefficient of friction drops out of the calculation of $\frac{\omega_2}{\omega_1}$.

Conserve angular momentum about bottommost point

$$(\frac25 MR^2)\omega_1 = (\frac25 MR^2) \omega_2 +MR\omega_2 R$$

$$\frac25 \omega_1=\frac75 \omega_2$$ so $$\frac{\omega_1}{\omega_2}=\frac72$$

Another method

Linear acceleration provided by frictional force=$f/M$

Angular retardation provided by friction=$\frac{f(R)}{\frac25 MR^2}$

When no slipping

$$\frac{f}{M} t=R(\omega_1- \frac{f(R)}{\frac25 MR^2}t)$$

$$f=\frac{RM}{t } \omega_1 -(\frac{ 5ft}{2M})M/t$$ $$\frac72 f=RM/t \omega_1$$

Hence $$\omega_2=\frac27 \omega_1$$

answered Nov 1, 2017 by (280 points)
edited May 30, 2018
Conservation of energy : $\frac12 I\omega_1^2 \ge \frac12 I\omega_2^2+\frac12 Mv_2^2$ where $v_2=R\omega_2$. Solving gives $\frac{\omega_2}{\omega_1} \le \sqrt{\frac27}$ as found by spacecube.
Your first method is now correct. Total AM about a point P is the AM about the CM plus the AM of the CM about P. I have deleted my first comment, which was incorrect : there is an external force (friction) but not an external torque, so angular momentum is conserved. Your 1st solution is simpler than the one I suggested. Your 2nd solution is essentially the one I suggested.