• Support PF! Buy your school textbooks, materials and every day products Here!

Hoop rolling in a pipe

  • Thread starter CAF123
  • Start date
  • #1
CAF123
Gold Member
2,895
88

Homework Statement


A hoop of mass ##m## and radius ##a## rolls without slipping inside a pipe of radius ##R## (the motion is 2D). Write down the kinetic and potential energy. Hence find the frequency of small oscillations about equilibrium.

Homework Equations


Moment of inertia of a hoop: I=Mr2
Rotational kinetic energy

The Attempt at a Solution


I envisaged the problem as a hoop rolling backwards and forwards in a pipe shaped like an arc of radius R.
Kinetic energy is sum of rotational and kinetic, so $$T = \frac{1}{2}mv^2 + \frac{1}{2} (ma^2) \left(\frac{v^2}{a^2}\right) = mv^2.$$

The path of the C.O.M of the hoop is arc shaped. Let ##\theta## be the angle between a vertical passing through the centre of the hoop and the C.O.M. Draw a horizontal at the base of the pipe. After a little time, the C.O.M is at angle ##\theta##. For small oscillations the increase in height of the C.O.M from its initial position (at a height a above the base of the pipe) is a + a(1-cosθ). So V = mg(a+a(1-cosθ)).

Is this correct? I am wondering if this can be solved more easily using the centre of momentum frame.
 

Answers and Replies

  • #2
3,812
92
V doesn't look correct. The CM of hoop is at a constant distance (R-a) from the centre of pipe. When you displace the hoop by angle ##\theta##, what is the distance of CM from the base?
 
  • #3
CAF123
Gold Member
2,895
88
Hi Pranav-Arora,
V doesn't look correct. The CM of hoop is at a constant distance (R-a) from the centre of pipe. When you displace the hoop by angle ##\theta##, what is the distance of CM from the base?
I believe that would be a + (R-a)(1-cosθ)
 
  • #4
3,812
92
I believe that would be a + (R-a)(1-cosθ)
Yes, looks right to me.
 
  • #5
CAF123
Gold Member
2,895
88
Yes, looks right to me.
Ok, so ##E = mv^2 + mg(a+(R-a)(1-\cos \theta))##. I think to find the freq of small oscillations, I should find a form ##\ddot{\theta}+ \omega^2 \theta = g##, where g is not a function of θ. Differentiating E would give me $$\ddot{\theta} + \left(\frac{g(R-a) - ga}{2a^2}\right)\theta = 0, $$from which I can extract ##\omega## and hence f.
 
  • #6
3,812
92
Ok, so ##E = mv^2 + mg(a+(R-a)(1-\cos \theta))##. I think to find the freq of small oscillations, I should find a form ##\ddot{\theta}+ \omega^2 \theta = g##, where g is not a function of θ. Differentiating E would give me $$\ddot{\theta} + \left(\frac{g(R-a) - ga}{2a^2}\right)\theta = 0, $$from which I can extract ##\omega## and hence f.
That doesn't look right.

Rewrite the potential energy as mg(R-(R-a)cosθ). Substitute in the energy equation and differentiate again.

What did you substitute for v?
 
  • #7
CAF123
Gold Member
2,895
88
That doesn't look right.
Could you please explain why? The equation is certainly dimensionally consistent.

What did you substitute for v?
Because of no slip, I said ##v_{COM} = a \dot{\theta}##.
 
  • #8
3,812
92
Because of no slip, I said ##v_{COM} = a \dot{\theta}##.
No, the hoop does not rotate about CM here, it rotates about the centre of pipe so ##v=(R-a)\dot{\theta}##.
 

Related Threads on Hoop rolling in a pipe

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
17
Views
7K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
568
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
931
Replies
2
Views
2K
Top