# Angular frequency of a mass performing SHM

• songoku
That is what guarantees that the sphere is rolling without slipping. There is no particular reason to presume that ##\dot \phi=0##, except that it makes the problem solvable. If you were given the velocity of the ball's CoM, you could solve for ##\dot \phi##. But it is true that ##\dot \phi=0## "at the start".

#### songoku

Homework Statement
A ball of mass m and radius r is put on a smooth surface having radius R (R > r). The ball is given a small displacement and then released so that it moves back and forth at the bottom of the surface. What is the angular frequency of the ball?
a. ##\omega = \sqrt{\frac{2g}{R}}##
b. ##\omega = \sqrt{\frac{g}{r}}##
c. ##\omega = \sqrt{\frac{g}{R}}##
d. ##\omega = \sqrt{\frac{g}{R-r}}##
e. ##\omega = \sqrt{\frac{g}{R+r}}##
Relevant Equations
Restoring force = m.a

Small angle approximation

##a=- \omega^{2} x##

When given a small displacement ##x##, the equation for m is:
(i) N sin θ = m.a where N is the normal force acting on the ball and θ is angle of the ball with respect to vertical.
(ii) N cos θ = m.g

So:
$$\tan \theta = \frac a g$$
$$\frac x R = \frac{\omega^{2} x}{g} \rightarrow \omega = \sqrt \frac{g}{R}$$

Is this correct? The size of m is not important?

Thanks

You have not defined r, but supposing it is the radius of the ball, what is the path of the ball’s center of mass?

songoku
Orodruin said:
You have not defined r, but supposing it is the radius of the ball, what is the path of the ball’s center of mass?
r has been defined by the question as the radius of the ball.

When given small displacement, the path of the ball's center of mass is arc of circle? Is this what you mean?

Thanks

songoku said:
the path of the ball's center of mass is arc of circle

songoku
haruspex said:
Ahh I see. The radius of the path is R - r so the answer should be (D).

Thank you very much Orodruin and haruspex

songoku said:
The size of m is not important?
Since ##\omega## can only depend on g, r, R and m, dimensional analysis shows that m is irrelevant.

songoku
haruspex said:
Since ##\omega## can only depend on g, r, R and m, dimensional analysis shows that m is irrelevant.
Indeed, and it is also very easy to see as ##m## is the only one of the five that contains non-trivial mass dimension.

More generally, dimensional analysis implies that
$$\omega = \sqrt{\frac{g}{R}} f(r/R).$$
In this case, the function ##f(x) = \sqrt{1/(1-x)}## cannot be deduced from dimensional analysis alone. The person constructing the question has been careful to make sure that all alternatives are dimensionally correct and therefore simply correspond to different ##f##.

On a side note, one thing that does bother me with this type of question is the assumption that the contact is frictionless so that the ball does not roll. This would be quite unrealistic in any experimental setup a student would make and this does affect the result. It is more realistic, particularly for low angular velocities, that the ball would roll without slipping. Edit: This is a nice extra credit problem. Assume that the ball is a homogeneous sphere and find out how the frequency changes when you instead make the assertion that it rolls without slipping.

Last edited:
berkeman, BvU and PeroK
I tried to use torque equation when solving the problem, yet the result differs significantly from the result when using translational method...
Does anyone know why?

Last edited:
Rikudo said:
I tried to use torque equation when solving the problem, yet the result differs significantly from the result when using translational method...
Does anyone know why?
Not unless you show us what you actually did.

Rikudo said:
I tried to use torque equation when solving the problem
Which problem, the one in post #1 (frictionless) that you already answered as D or the frictional variant @Orodruin suggests in post #7?

The problem of post #1.
$$-mgsin\theta(R-r)= I\ddot\theta$$
Since ##\theta## is small, and ##I = 2mr²/5 + m(R-r)²##:
$$-mg\theta(R-r)=[2mr²/5 + m(R-r)²] \ddot\theta$$
From the above equation, it is clear that it is different from the translational:
$$-mgsin\theta = m\ddot\theta (R-r)$$

Rikudo said:
$$-mg\theta(R-r)=[2mr²/5 + m(R-r)²] \ddot\theta$$
This equation is not set up correctly. It says that when the CoM of the sphere has an angular displacement of ##\theta##, the rolling sphere rotates by the same ##\theta## about its center. It does not. It rotates by a different angle ##\phi##. In other words, the orbital angular momentum of the sphere is ##L_{\text{orb}}=m(R-r)^2\dot {\theta}## and the spin angular momentum is ##L_{\text{spin}}=\frac{2}{5}mr^2\dot {\phi}##. You need to find the constraint relation between ##\theta## and ##\phi.##

Last edited:
Rikudo
Rikudo said:
The problem of post #1.
$$-mgsin\theta(R-r)= I\ddot\theta$$
Since ##\theta## is small, and ##I = 2mr²/5 + m(R-r)²##:
$$-mg\theta(R-r)=[2mr²/5 + m(R-r)²] \ddot\theta$$
From the above equation, it is clear that it is different from the translational:
$$-mgsin\theta = m\ddot\theta (R-r)$$
In the problem of post #1 there is no friction, so no torque about the ball's centre. It will not rotate. The moment of inertia is irrelevant.

Rikudo
kuruman said:
This equation is not set up correctly. It says that when the CoM of the sphere has an angular displacement of ##\theta##, the rolling sphere rotates by the same ##\theta## about its center. It does not. It rotates by a different angle ##\phi##. In other words, the orbital angular momentum of the sphere is ##L_{\text{orb}}=m(R-r)^2\dot {\theta}## and the spin angular momentum is ##L_{\text{spin}}=\frac{2}{5}mr^2\dot {\phi}##. You need to find the constraint relation between ##\theta## and ##\phi.##
Ah. Sinc it is not spinning, ##\dot \phi## is 0. Is this true?

Rikudo said:
Ah. Sinc it is not spinning, ##\dot \phi## is 0. Is this true?
Correct. I thought you were doing the spinning problem.

Rikudo said:
Ah. Sinc it is not spinning, ##\dot \phi## is 0. Is this true?
Not quite. What you can say, and what matters, is that ##\ddot \phi=0##.