# SHM of a rigid body

1. Mar 26, 2016

### erisedk

1. The problem statement, all variables and given/known data
A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end. A thin disc of mass ‘M’ and radius ‘R’ (<L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached: (case A) The disc is not free to rotate about its centre and (case B) the disc is free to rotate about its centre.
The rod disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true?

(A) Restoring torque in case A = Restoring torque in case B
(B) restoring torque in case A < Restoring torque in case B
(C) Angular frequency for case A > Angular frequency for case B.
(D) Angular frequency for case A < Angular frequency for case B.

2. Relevant equations
$\vec{τ} = I\vec{α}$ ............... Eqn (i)
$\vec{α} = \dfrac{\vec{a}}{r}$ .................Eqn (ii)
$\vec{α} = ω^2θ$ ................Eqn (iii)
T for a physical pendulum = $2π\sqrt{\dfrac{I}{mgd}}$ where I is the moment of inertia about the point of suspension and d is the distance of the centre of mass of the rigid body from the point of suspension. (Eqn (iv))

3. The attempt at a solution
To check options C and D,
(case A) I = mL2/3 + ML2
(case B) I = mL2/3 + ML2 + MR2/2

Since I in (case A) is smaller than I in (case B)
According to Eqn (iv) of relevant equations, ωA > ωB (as ω = 2π/T) i.e. (C)

Restoring torque (i.e to check options A and B):
$\vec{α} = ω^2θ$
Since both are displaced from same position θ is the same. So α depends on ω.
Which means αA > αB and IA < IB
Also, I will get cancelled out in the torque expression because αI = ω2θI and ω2 is inversely proportional to I.
So, torques will be equal. Hence (A) will hold.
However, the answer is (A) and (D).

2. Mar 26, 2016

### TSny

Case A is the case where the disk is not free to rotate about its center. Therefore, in this case the disk will be forced to rotate along with the rod. In case B the disk is free to rotate about its center, so the disk will not rotate with the rod.

3. Mar 26, 2016

### erisedk

Yeah, that's what I've tried to do. That's why I haven't considered the disc's MOI in case A because in case A it simply behaves like a mass attached at the end of a rod.

4. Mar 26, 2016

### TSny

See the figure below which represents my interpretation of the wording of the problem. Note that in case A the disk rotates while in B it does not.

#### Attached Files:

• ###### disk pendulum 3.png
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5. Mar 26, 2016

### erisedk

Oh! Got it. So, the MOI will just be the reverse of what I've written. Thank you so much! :)

6. Mar 26, 2016

### Vibhor

Hello TSny,

Please clarify the following points .

1) In your figure in case A , the angular speeds of points 'a' and 'b' would be same and that in case B ,would be different . Right ??

2) In case B , even though the disk does not rotate about its center but it (disk) does rotate about the fixed end of the rod (By definition of rotation of a body about a point). Different points on the disk moving with different speeds/angular speeds . Right ??

Last edited: Mar 27, 2016
7. Mar 27, 2016

### TSny

Yes, for case A, points 'a' and 'b' would have the same angular speed about the fixed end of the rod.

For case B the points would have different angular speed about the fixed end of the rod. That is, if you used polar coordinates $(r, \theta)$ with origin at the fixed end of the rod, then $\dot{\theta}$ for the points 'a' and 'b' would be different (at the same instant of time). In this sense, if a particle moves in the x-y plane along the straight line y = 5, it would have an angular speed $\dot{\theta}$. I'm not sure if I would say that the particle is "rotating" about the origin.

For case B, different points on the disk would have different angular speeds $\dot{\theta}$ relative to the fixed end of the rod. Nevertheless, the disk (as a rigid body) would be said to be in pure translation. More specifically, it would be in pure curvilinear translation. See figures 51.2 and 15.4 here: http://highered.mheducation.com/sites/007230491x/student_view0/chapter15/chapter_overview.html

8. Mar 27, 2016

### Vibhor

Thanks .

In case B ,the MI is calculated by assuming mass of the disk to be present at the center i.e MI in case B = mL2/3 + ML2 .

Could you please explain why do we assume disk to be a point mass at its center for calculating MI about the fixed end of rod ( in case the disk is free to rotate about its center)??

9. Mar 27, 2016

### TSny

There is a general theorem about angular momentum that holds for any system of particles. The theorem says that the angular momentum of a system relative to some fixed point can be written as the sum of two parts. The first part equals the angular momentum that would be obtained if you concentrated the total mass of the system at the center of mass and treated the total mass as a particle moving with the center of mass. The second part equals the angular momentum of all of the particles with respect to the moving center of mass of the system. Section 5.1.4 of this link gives the proof: http://www.damtp.cam.ac.uk/user/stcs/courses/dynamics/lecturenotes/section5.pdf

Treating the disk as a system of particles, the angular momentum of the disk relative to the upper end of the rod will equal the sum of two terms:
Term 1: the angular momentum due to treating the disk as a point particle at the center of the disk
Term 2: the angular momentum of the disk relative to the center of the disk.

In case B, term 2 is zero. Term 1 can be written as $ML^2 \omega$ where $\omega$ is the angular velocity of the rod and $M$ is the mass of the disk.

10. Mar 27, 2016

Thanks a lot