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

Rolling disk with point mass on edge

  • Thread starter vladimir69
  • Start date
  • #1
130
0

Homework Statement


A point mass m is attached to the rim of an otherwise uniform solid disk of mass M and radius R (in the diagram mass m is in contact with the ground). The disk is rolled slightly away from its equilibrium position and released. It rolls back and forth without slipping. Show that the period of this motion is given by
[tex] T=2\pi \sqrt{\frac{3MR}{2mg}}[/tex]


Homework Equations


[tex]I=\frac{1}{2}MR^2[/tex]
[tex]I \frac{d^2 \theta}{dt^2}=-\kappa \theta[/tex]
[tex]\omega=\sqrt{\frac{\kappa}{I}}[/tex]
[tex]T=\frac{2\pi}{\omega}[/tex]
[tex]\alpha = \frac{d^2\theta}{dt^2}[/tex]

The Attempt at a Solution


My guess would have been
[tex]I=\frac{1}{2}MR^2 + mR^2[/tex]
and then
[tex]I\alpha = -mgR\sin\theta \approx -mgR\theta[/tex]
now we have kappa and I so we can pop that in to solve for T which gives a big mess. This problem is a bit tricky for me as I am trying to look for a stationary axis to take torques about and not sure about how to apply the parallel axis theorem with the point mass (do I even need to use it?). I tried calculating the centre of mass of the disk just for kicks and it turned out to be a point 2/3R from the centre (using Centre of Mass = [itex]\int r dm[/itex]) but it should be in the centre...thats another story.

It seems there must be a pretty easy way to solve it but I cant see it.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
A point mass m is attached to the rim of an otherwise uniform solid disk of mass M and radius R (in the diagram mass m is in contact with the ground). The disk is rolled slightly away from its equilibrium position and released. It rolls back and forth without slipping. Show that the period of this motion is …

… I am trying to look for a stationary axis to take torques about …

It seems there must be a pretty easy way to solve it but I cant see it.
Hi vladimir69! :smile:

Hint: conservation of energy :wink:
 
  • #3
130
0
Thanks for the reply
Not sure how using conservation of energy will allow me to find the period of oscillations for this problem. Not 100% sure how to apply conservation of energy for this problem either.
 
  • #4
130
0
I am tempted to work it out by trying to find a way so that (analogous to a pendulum)
[tex]\frac{1}{2}MR^2\alpha=-\frac{1}{3}mgR\theta[/tex]
because from the above equation you can get omega and then immediately go to the period T using [itex]T=2\pi\frac{1}{\omega}[/itex]

edit: not sure why my latex is not showing correctly, it is displaying stuff I made in previous posts.
 
Last edited:
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
249
I am tempted to work it out by trying to find a way so that (analogous to a pendulum)
[tex]\frac{1}{2}MR^2\alpha =-\frac{1}{3}mgR\theta[/tex]
because from the above equation you can get omega and then immediately go to the period T using [itex]T=2\pi\frac{1}{\omega}[/itex]
vladimir69 said:
i am still working on it but cant proceed
my brain is still stuck in thinking of pendulums and not of conservation of energy
[tex]
\frac{1}{2}MR^2\alpha=-mgR\theta
[/tex]
thats my pendulum equation for this system but its missing a factor of a 1/3

hope you can help
thanks tiny tim
Hi vladimir69! Thanks for the PM. :smile:

I think conservation of energy is the easiest way …

the KE of the disc is its linear KE (1/2 Mv2) plus its rotational KE (1/2 Iθ'2) …

(and add the KE of the point mass, but you should find that it's second-order and can be ignored :wink:)

and the PE is mgR(1 - cosθ), ~ 1/2 mgRθ2

then write v as a function of R and θ, ignore anything of second order in θ,

and the equation KE + PE = constant should give you

(something)θ'2 + (something else)θ2 = constant,

which you can then solve. :smile:
 

Related Threads for: Rolling disk with point mass on edge

Replies
3
Views
642
  • Last Post
Replies
3
Views
309
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
4
Views
900
Replies
24
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
15K
Top