# An overall question about rotational dynamics

1. Dec 6, 2011

### yttuncel

1. The problem statement, all variables and given/known data

A dumbbell is made of a rod length L and mass M, and 2 spheres of radius r and mass m (see figure). The rod is attached at a distance L/3 from the left end to a
rotational axis A. The dumbbell is let go under gravity and makes a rotational motion under gravity around the rotational axis A. (Known: g, M, m, L, r)
a) Make a detailed analysis of the problem
b) Find the distance between the center of mass of the system and the rotational axis A.
c) Find the moment of inertia (I) of the system about A.
d) What is the maximum angular velocity achieved by the system?
e) What are the maximum speed of each of the spheres?

2. Relevant equations

I=MR2
xcm=(Ʃmx)/M

3. The attempt at a solution

b) I dont get what it meant, shall i say L/6 directly or anything else? Compute center of mass from the formula? But how?

c) I found an answer from parallel axis theorem. That is:
I=2/5mr2+m(r+L/3)2+2/5mr2+m(r+2L/3)2+1/12ML2+1/36ML2

Currently I am at the d) part, please comment and help :)

2. Dec 6, 2011

### Staff: Mentor

If you have to show your work for part (b) then you can argue that the dumbbell is symmetric about its center of mass, so the COM must be located at L/2 on the rod. Then your result follows given the position of point A.

For part (c) you can probably simplify that expression a bit more to to make it easier for the marker to check it against a standard answer.

For part (d) you'll have to determine what forces (torques) are acting and what the resulting kinematic equation is going to be. Draw a free body diagram.

3. Dec 6, 2011

### yttuncel

Ok, I got it. For part d) I calculated net torque on the system.
τ=-m(r+L/3)+M(L/6)+m(r+2L/3)
τ=L/6(2m+M)
Now ?
τ=I*α
How will I relate to velocity? We do not know Δt

4. Dec 6, 2011

### Staff: Mentor

Look at the system and see if you can determine at what position the velocity should be maximum. You can also consider conservation of energy.

5. Dec 7, 2011

### yttuncel

Vel. should be max when the right hand sphere is at the bottom.
Right sphere's velocity = V1 = w(2L/3+r)
Left sphere's velocity = V2 = w(L/3+r)
Rod's center of mass's velocity = V3 = w(L/6)

ΔK+ΔU=0
ΔK=1/2m(V12+V22)+1/2MV32
ΔU=mg(L/3+r)-mg(2L/3+r)-MgL/3

Anything else I should do?

6. Dec 7, 2011

### Staff: Mentor

I don't see how you've found a value for ω, so you can't determine the various velocities.

Why not calculate the change in gravitational potential energy between the starting position and the position that you've identified as corresponding to the highest angular velocity? (Draw all the components for both positions and calculate the change in height for each. Sum the changes in PE)

Then you can use the expression for angular KE and your formula for the ensemble's moment of inertia to find the angular velocity of the system.

Last edited: Dec 7, 2011
7. Dec 7, 2011

### yttuncel

ΔU=mg(L/3+r)-mg(2L/3+r)-MgL/3 isnt this the change in gravitational potential energy?
So i will equalize 1/2Iw^2 with the one above right? And -I- will be the one in part c right?

8. Dec 7, 2011

### Staff: Mentor

The change in PE associated with the rod doesn't look quite right. Can you expand on your calculation a bit?

9. Dec 7, 2011

### yttuncel

Oops you are right. It is
ΔU=mg(r+L/3)-mg(r+2L/3)-MgL/6

10. Dec 8, 2011

### Staff: Mentor

Good. You can probably combine the two mg terms, too, while you're at it. Just to be neat & tidy