# Homework Help: Rigid body rotation

1. Nov 25, 2008

### wolf party

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

imagine a thin length of wood supported by two supports, one at either end. One support dissapears and the other turns into a pivot (as gravity acts on the wood). Show that the load supported by the pivot is Mg/4.

3. The attempt at a solution

i dont know where to start due to my confusion on forces with rotation. i know the plank is acted on by gravity at its CM, and its moment of inertia about its CM = I=1/12*M*l^2.
I know there is a reacton force at the Pivot, and this is the force i need to calculate, but i dont know what to consider first, a nudge in the right direction would be helpful!

2. Nov 25, 2008

### Staff: Mentor

Start by figuring out the acceleration of the center of mass. Hint: Apply Newton's 2nd law to the rotation.

3. Nov 25, 2008

### wolf party

F=ma -> TAU = I*ALPHA

TAU = (1/12*M*l*l)*ALPHA

so ALPHA = TAU/(1/12*M*l*l)

so a = (TAU/(1/12*M*l*l)) * (0.5*l) ? where a=ALPHA*(0.5l)

4. Nov 25, 2008

### Staff: Mentor

(1) It pivots about one end, not the center of mass, so correct the rotational inertia.
(2) What's the torque about the pivot?

5. Nov 25, 2008

### wolf party

TAU = (1/3*M*L*L)*ALPHA

the torque about the pivot is mgl/2 ?

6. Nov 25, 2008

### Staff: Mentor

Good. Keep going.

7. Nov 25, 2008

### wolf party

so the acceleration due to gravity is a = 3*TAU/2*ML at the CM
and the torque about the pivot is mgl/2

so m*a = TAU*(1/2*l) ?

8. Nov 25, 2008

### Staff: Mentor

Good. So what's the acceleration of the center of mass? (Don't call it "acceleration due to gravity".)

Once you have the acceleration of the center of mass, apply Newton's 2nd law for translation. What forces act on the wood?

9. Nov 25, 2008

### wolf party

is the acceleration ofthe CM just 3TAU/2mL again
the reacton force at the pivot and mg at the CM are the forces on the wood

R = 0 because mgl/2 (where l = 0) = 0

therfore ma=mg ?

10. Nov 25, 2008

### wolf party

acceleration at CM = 3/4*g

11. Nov 25, 2008

### Staff: Mentor

But you know the torque, so eliminate Tau from this expression.
Good.
No. Apply Newton's 2nd law. What's net force? The acceleration?

12. Nov 25, 2008

### Staff: Mentor

Good!

13. Nov 25, 2008

### wolf party

the force at the CM=3/4*mg

Net force at pivot = force at CM ?

14. Nov 25, 2008

### Staff: Mentor

That's the "ma" part of "Fnet = ma". But what's Fnet?
No. Two forces act on the wood (you named them earlier). Combine them to get the net force.

15. Nov 25, 2008

### wolf party

Fnet = mg + mgl/2
?

16. Nov 25, 2008

### wolf party

FNet = N - mg = 3/4mg

therfore N = 1/4mg ?

17. Nov 25, 2008

### Staff: Mentor

Careful. As written, this implies that N = 7/4mg. But the acceleration is downward, thus negative: N - mg = -3/4mg.

Yep.

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