- #1

jgens

Gold Member

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## 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 d

*F*be the force exerted by the pivot on a small segment of mass d

*M*at a radial distance

*r*from the pivot. This means that ...

d

*F*=

*w*

^{2}

*r*d

*M*

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*= d

*M*/d

*r*=

*M/l*. Using this information in the above equation, we get ...

d

*F*= k

*w*

^{2}

*r*d

*r*

Integrating this gives ...

[tex]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}[/tex]

Is this right so far?