Angular frequency of a mass performing SHM

In summary: 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".
  • #1
songoku
2,365
347
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##
1651123246249.png

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
 
Physics news on Phys.org
  • #2
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?
 
  • Like
Likes songoku
  • #3
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
 
  • #4
songoku said:
the path of the ball's center of mass is arc of circle
Of what radius?
 
  • Like
Likes songoku
  • #5
haruspex said:
Of what radius?
Ahh I see. The radius of the path is R - r so the answer should be (D).

Thank you very much Orodruin and haruspex
 
  • #6
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.
 
  • Like
Likes songoku
  • #7
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:
  • Like
Likes berkeman, BvU and PeroK
  • #8
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:
  • #9
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.
 
  • #10
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?
 
  • #11
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:
$$-mgsin\theta = m\ddot\theta (R-r)$$
 
  • #12
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:
  • Like
Likes Rikudo
  • #13
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.
 
  • Like
Likes Rikudo
  • #14
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?
 
  • #15
Rikudo said:
Ah. Sinc it is not spinning, ##\dot \phi## is 0. Is this true?
Correct. I thought you were doing the spinning problem.
 
  • #16
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##.
 

FAQ: Angular frequency of a mass performing SHM

What is the definition of angular frequency in the context of simple harmonic motion?

The angular frequency of a mass performing simple harmonic motion (SHM) is defined as the rate at which the mass oscillates back and forth in a circular motion. It is measured in radians per second and is denoted by the symbol ω.

How is the angular frequency related to the period of SHM?

The angular frequency and the period of SHM are inversely proportional to each other. This means that as the angular frequency increases, the period decreases, and vice versa. The relationship between the two can be expressed as T = 2π/ω, where T is the period and ω is the angular frequency.

What is the formula for calculating the angular frequency of a mass in SHM?

The formula for calculating the angular frequency of a mass in SHM is ω = √(k/m), where k is the spring constant and m is the mass of the object. This formula can also be written as ω = 2πf, where f is the frequency of oscillation in hertz.

How does changing the spring constant or mass affect the angular frequency of SHM?

Changing the spring constant or mass of an object in SHM will directly affect the angular frequency. A higher spring constant or a lower mass will result in a higher angular frequency, and a lower spring constant or a higher mass will result in a lower angular frequency. This is because the angular frequency is dependent on the stiffness of the spring and the mass of the object.

Can the angular frequency of SHM be negative?

No, the angular frequency of SHM cannot be negative. It is always a positive value, as it represents the speed and direction of the oscillation of the mass. A negative angular frequency would indicate that the mass is moving in the opposite direction of the oscillation, which is not possible in SHM.

Back
Top