An overall question about rotational dynamics

[yttuncel]

Homework Statement

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?

http://imageupload.org/thumb/thumb_145674.jpg

Homework Equations

I=MR²
x_cm=(Ʃmx)/M

The Attempt at a Solution

b) I don't 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 :)

 

[gneill]

Homework Statement

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?

http://imageupload.org/thumb/thumb_145674.jpg

Homework Equations

I=MR²
x_cm=(Ʃmx)/M

The Attempt at a Solution

b) I don't 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 :)

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.

 

[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

 

[gneill]

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

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.

 

[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?

 

[gneill]

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?

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.

 

[yttuncel]

ΔU=mg(L/3+r)-mg(2L/3+r)-MgL/3 isn't 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?

 

[gneill]

ΔU=mg(L/3+r)-mg(2L/3+r)-MgL/3 isn't 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?

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

 

[yttuncel]

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

 

[gneill]

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

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