Falling stick

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A thin, uniform stick of length 2.1 m and mass 4.2 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. Remember that the moment of inertia for a stick of mass m and length L about its end is (1/3)m L2.

Also, using conservation of energy, it can be shown that the square of the angular speed as a function of angle is given by:

ω2 = 3 g sin(θ)/L

with θ the angle measured clockwise from horizontal and L the length of the stick.

What is the magnitude of the angular acceleration of the stick?
What is the magnitude of the tangential acceleration of the center of mass of the stick?
What is the magnitude of the centripetal acceleration of the center of mass of the stick?
What is the magnitude of the total acceleration of the center of mass of the stick?

I dunno know even where to begin, I am having a hard time with this subject area. It would be nice to get full drawn out answers. Thanks.
 

Answers and Replies

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The rules of the forum mean that you have to have a go at the question before anyone can correct your answer.

To help you start, I like to compare angular motion to ordinary linear motion, but:-

Force becomes Torque (where Torque = force *perp. dist to pivot)
Mass m becomes moment of inertia I (youve got I from the question)
Initial velocity u becomes initial angular velocity w0[omega subscript 0] (u=r*w0)
Final velocity v becomes final ang. vel. w1
Acceleration a becomes ang. acc. alpha (a=r*alpha)
Time is still time t
Displacement s becomes ang. displacement theta. (s=r*theta)

You can then 'translate' all your linear physics equations to angular ones

e.g v=u+at becomes w1=w0+alpha*t
kinetic energy 1/2mv^2 becomes 1/2 I w^2
F=ma becomes Torque=I *alpha
etc.

Draw your stick at an angle theta to the horizon and see what you can come up with.
 
Last edited:

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