A stick with a mass of 0.164 kg and a length of 1.00 m is pivoted about one end

Question

A stick with a mass of 0.164 kg and a length of 1.00 m 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.
a)As it swings through the vertical, calculatethe change in gravitational potential energy that has occurred.
b)As it swings through the vertical, calculate the angular speed of the stick.
c)As it swings through the vertical, calculate the linear speed of the end of the stick opposite the axis.
d)Find the ratio of the speed of a particle that has fallen a distance of 1.00 m, starting from rest, to the speed from part (C).

Explanation / Answer

a) change in PE = final - initial final is at bottom so = 0 initial = mgh therefore: 0 - mgh = -(.17)(9.8)(.5) = - .833 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 = .833 1/6(.17)(1^2)w^2 = .833 w = sqrt(.833*6/.17) w = 5.42218 c) w = V/R R = 1m therefore: w = V 5.42218 = V

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