Rigid Body Rotation: Calculating Load Supported by Pivot

[wolf party]

Homework Statement

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.

The Attempt at a Solution

i don't 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 don't know what to consider first, a nudge in the right direction would be helpful!

 

[Doc Al]

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

 

[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)

 

[Doc Al]

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

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

 

[wolf party]

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

the torque about the pivot is mgl/2 ?

 

[Doc Al]

Good. Keep going.

 

[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) ?

 

[Doc Al]

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

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?

 

[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 ?

 

[wolf party]

acceleration at CM = 3/4*g

 

[Doc Al]

is the acceleration ofthe CM just 3TAU/2mL again

But you know the torque, so eliminate Tau from this expression.

the reacton force at the pivot and mg at the CM are the forces on the wood

Good.

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

therfore ma=mg ?

No. Apply Newton's 2nd law. What's net force? The acceleration?

 

[Doc Al]

acceleration at CM = 3/4*g

Good!

 

[wolf party]

the force at the CM=3/4*mg

Net force at pivot = force at CM ?

 

[Doc Al]

the force at the CM=3/4*mg

That's the "ma" part of "F_net = ma". But what's F_net?

Net force at pivot = force at CM ?

No. Two forces act on the wood (you named them earlier). Combine them to get the net force.

 

[wolf party]

Fnet = mg + mgl/2
?

 

[wolf party]

FNet = N - mg = 3/4mg

therfore N = 1/4mg ?

 

[Doc Al]

FNet = N - mg = 3/4mg

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

therfore N = 1/4mg ?

Yep.