A meter stick with a mass of 0.159kg is pivoted about one end so it can rotate w

Question

A meter stick with a mass of 0.159kg is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate:

A.) As it swings through the vertical, calculate the change in gravitational potential energy that has occurred.

B.) Calculate the angular speed of the stick.

C.) Calculate the linear speed of the end of the stick opposite the axis.

D.) Compare the answer in part C to the speed of a particle that has fallen 1.00m , starting from rest.

Explanation / Answer

a) change in PE = final - initial
final is at bottom so = 0
initial = mgh
therefore:

0 - mgh = -(.159)(9.8)(.5) --length (1m so1/2=0.5)( Compared to where it started, the center of mass of the stick is now half the length of the stick lower. )

= -0.779J

b) Work = KE + PE
here work is 0 (no friction)
KE = final - intial
final = 1/2Iw^2
initial = 0
in this case I = 1/3ML^2
therefore:

1/6ML^2w^2 = .779
1/6(.159)(1^2)w^2 = .779
w = sqrt(.779*6/.159)
w = 4.8rad/sec

c)For a rotating object:
linear velocity = angular velocity x radius

R = 1m

Linear velocity=4.8*1

Linear velocity==4.8m/sec

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