# How does rotational energy transfer to linear energy?

So I have recently started looking into moments of inertia, and all that stuff. I have come to a question which has a plane inclined at some angle theta and a sphere at the peak. The G.P.E at the top is mgh. The plane is frictionless.The book then states by conservation of energy $mgh = 1/2(mv^2) + \text{rotational KE from the sphere}$.

Firstly I don't understand how if the ball is initially rotating how it gains rotational energy from the top of the plane to the bottom, since there is no friction to provide torque. (Or how released from rest how it begins to rotate) Similarly is it the case that mass is always treated as if it is all concentrated at the center of mass, because intuitively I would have thought that the individual mass particles making up the sphere might have provided some torque?

Sorry if this is a bit of a basic question but I've only just started the topic.

Any help would be much appreciated!

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• Are you sure there is no friction between plane and sphere? – Aretino Jul 30 '15 at 16:24
• I've voted to move this to physics.SE – Zev Chonoles Jul 30 '15 at 17:18
• I have also suggested that a moderator moves the question to the physics SE community. – wltrup Jul 30 '15 at 17:30
• It's an applied mathematics question! – David Quinn Jul 30 '15 at 18:29

Firstly I don't understand how if the ball is initially rotating how it gains rotational energy from the top of the plane to the bottom, since there is no friction to provide torque.

Your understanding is mostly correct. We can choose any point on the object to measure rotation around; let us choose the center of the ball. Both the gravitational force and the normal force point through this center and cannot rotate the ball.

However, "no friction" could also mean "it rolls without slipping and there is negligible rolling friction." In that case, there is static friction providing torque, it just does not do work on the ball. If this is the case, the explanation will involve the moment of inertia $I$ as:$$m~g~h = \frac 12 ~m ~v^2 + \frac 12 ~I~ \omega^2$$and the fact that $\omega = \pm v/R$ will then be substituted to get $$2~g~h = \left(1 + \frac I{m ~R^2}\right) ~ v^2$$For a hollow sphere this ratio $I/mR^2$ is $2/3$; for a solid, uniform ball it is $2/5$ since more of it is closer to the center and therefore easier to rotate.

Similarly is it the case that mass is always treated as if it is all concentrated at the center of mass, because intuitively I would have thought that the individual mass particles making up the sphere might have provided some torque?

For any rigid body we can decouple the dynamics into two parts. The first part is the dynamics of the center of mass, which can be treated as having a mass equal to the total mass of the rigid body and as having forces equal to the sum of all external forces on the rigid body (internal ones tend to cancel due to Newton's third law, so you can include them too if you're careful to always include them in pairs). If the net force on particle $i$ is Newton's law, $$m_i \frac{d\vec v_i}{dt} = \vec F_{i},$$ then we can find the center of mass with the formula $\vec R = \sum_i \frac{m_i}{M} ~ \vec r_i$ where $M = \sum_i m_i.$ This in turn moves with a velocity $\vec V = \sum_i \frac{m_i}{M} ~ \vec v_i$, so then $M ~ \frac{d\vec V}{dt} = \sum_i \vec F_i$. If you're paying very close attention you can even see that there's no absolute meaning to $M$, except that it keeps the center of mass $\vec R$ somewhere which is an "average" of all the positions of the little masses, which is nice (but not essential) for understanding the meaning of $\vec V$.

The second part is the rotation of the body about the center of mass. This is a little more complicated. Fortunately we have a nice theorem about the rotation group SO(3): every rotation is about some axis, also in 3D space, which it leaves stationary. We can therefore identify rotation with a vector $\vec \omega$, with each particle involved in the rotation having position $\vec r_i$ and velocity $\vec v_i = \vec \omega \times \vec r_i$, which is called the "cross product" of the two vectors. We can sum all of the details up by an "angular momentum" about a fixed point $\vec r_0$, $\vec L = \sum_i m_i~(\vec r_i - \vec r_0)~\times~\vec v_i$. Since $\vec v_i \times \vec v_i = 0$ the sum of Newton's rotational laws can be written, $$\frac {d\vec L}{dt} = M \vec V \times \frac{d\vec r_0}{dt} + \sum_i m_i ~(\vec r_i - \vec r_0) \times \frac{d\vec v_i}{dt}$$ So when $\vec r_0 = \vec R$ and we get a $\vec V \times \vec V = 0$ term, this is simply$$\frac{d\vec L}{dt} = = \sum_i (\vec r_i - \vec R) \times \vec F_i = \sum_i \vec \tau_i.$$ So, we sum together these "torques" about the center of mass and they tell us how this "angular momentum" of the object changes. Then it turns out that the angular momentum always exists in a fixed relationship to $\vec \omega$ given by the "moment of inertia tensor" $\mathbf I$ as $\vec L = \mathbf I \cdot \vec \omega$, so any change in the angular momentum creates the corresponding change in the angular velocity vector.

• Your explanation notwithstanding, your statement However, "no friction" could also mean "it rolls without slipping and there is negligible rolling friction." is changing the problem. "No friction" can't mean "some friction." If the person who wrote the problem meant what you suggested, then that person is ill-informed or deceptive. – Bill N Jul 31 '15 at 2:35
• @BillN I don't know anyone who's made it through a few college physics classes without running into some situation where a professor asked the wrong question. I think I've only had one course that was totally devoid of it, and that was only because it was taught by this guy and I'm pretty convinced that he has fielded every undergraduate physics question known to mankind. – CR Drost Jul 31 '15 at 3:06

If the plane if frictionless, then any initial angular velocity the sphere may have been given will remain constant. This is because there is no torque on the sphere. The rotational KE will therefore remain constant throughout the motion.

The KE of a rolling sphere is the linear KE of the centre of mass + the rotational KE about the centre of mass