Rotational Dynamics of a plank of mass

[jgens]

Homework Statement

A thin plank of mass M and length l is pivoted at one end. The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.

Homework Equations

N/A

The Attempt at a Solution

So, here's what I worked out ...

Reasoning:
I figured that when the plank is horizontal there are two forces acting on it: The graviational force and the force of the pivot on the plank (which is a radial force). Since the gravitational force has no component in the radial direction when the plank is horizontal, the force required to keep the plank in uniform circular motion must supplied by the force of the pivot on the plank. Using Newton's Third Law, we see that the force of the plank on the pivot is equal in magnitude and opposite in direction to the force of the pivot on the plank. Therefore, we just need to calculate the force of the pivot on the plank (the centripetal force needed to keep the plank in uniform circular motion) in order to find the desired quantity. Denote this force by F.

Calculations:
Let dF be the force exerted by the pivot on a small segment of mass dM at a radial distance r from the pivot. This means that ...

dF = w²r dM

where w is angular velocity of the rod when it's horizontal. Assuming that the mass density of the rod is uniform, we see that k = dM/dr = M/l. Using this information in the above equation, we get ...

dF = kw²r dr

Integrating this gives ...

F = \int_0^l k\omega^2r \, dr = k\omega^2\int_0^lr \, dr = \frac{k\omega^2l^2}{2} = \frac{M\omega^2l}{2}


Is this right so far?

 

[jgens]

Calculations cont:
We can find the net torque acting on the plank if we treat it as if all of the mass were concentrated at the center of mass. Therefore, this torque is given by ...

\tau = I\alpha = \frac{ML^2}{3}\alpha = \frac{Mgl}{2}\sin{(\theta)}

This can be written ...

\alpha = \ddot{\theta} = \frac{3g}{2l}\sin{(\theta)}

Therefore,

\dot{\theta}\ddot{\theta} = \frac{3g}{2l}\sin{(\theta)}\dot{\theta}

Integrating this, we see that ...

\frac{\dot{\theta}^2}{2} \bigg{|}_0^{\omega} = \frac{3g}{2l}\int_{\pi/3}^{\pi/2}\cos{\theta}\,d\theta

Solving this equation, we get ...

\frac{\omega^2}{2} = -\frac{3g}{2l}\cos{(\theta)}\bigg{|}_{\pi/3}^{\pi/2} = \frac{3g}{4l}

Is this right in the context of the problem?

 

[tiny-tim]

hi jgens!

A thin plank of mass M and length l is pivoted at one end. The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.
…
Reasoning:
I figured that when the plank is horizontal there are two forces acting on it: The graviational force and the force of the pivot on the plank (which is a radial force). Since the gravitational force has no component in the radial direction when the plank is horizontal, the force required to keep the plank in uniform circular motion must supplied by the force of the pivot on the plank. Using Newton's Third Law, we see that the force of the plank on the pivot is equal in magnitude and opposite in direction to the force of the pivot on the plank. Therefore, we just need to calculate the force of the pivot on the plank (the centripetal force needed to keep the plank in uniform circular motion) …

that's really good!

unfortunately you then left the path of reason and went berserk with equations about torque

go back to your reasoned argument … horizontal force = mass times centripetal acceleration of the centre of mass …

so get the acceleration from the angular velocity, and get the angular velocity from conservation of energy!

(btw, you didn't need that integration in your first post … if you look at ∫ r dr, you can see it's bound to always give you the position of the centre of mass, in this case at L/2! )

 

[jgens]

Okay, so for conservation of energy, we can express everything in terms of translation and rotation of the center of mass. Initially, all of the energy is potential energy, and the potential energy at the CM is ...

Mgh = \frac{Mgl\cos{\theta}}{2} = \frac{Mgl\cos{\pi/6}}{2} = \frac{Mgl\sqrt{3}}{4}

By the Law of Conservation of Mechanical Energy, this is equal to the rotational and translational kinetic energy of the CM. Since the CM isn't rotating, this gives us the equation

\frac{Mgl\sqrt{3}}{4} = \frac{1}{2}MV^2 = \frac{1}{8}M\omega^2l^2

Solving for \omega^2 yields,

2\sqrt{3}\frac{g}{l}


This doesn't look right though, so I think that I've done something wrong. By the way, what doesn't work about my argument with torque?

 

[tiny-tim]

Since the CM isn't rotating …

uhh? that doesn't mean anything!

the rod is rotating, so you do need to add the rotational KE

(and btw, either you used the wrong angle, or you used cos instead of sin)

By the way, what doesn't work about my argument with torque?

it's probably ok, but i couldn't be bothered to check it!

try again!

(and now I'm off to bed :zzz:)​

 

[jgens]

Okay! Now I understand you. And going through the calculations it looks like it should give me the same result as my argument with torque. That's a relief.

Thanks!