A uniform, slender rod of mass 0.52 kg and length 0.81 m is initially tilted at

Question

A uniform, slender rod of mass 0.52 kg and length 0.81 m is initially tilted at an angle 40 degrees above the horizontal with one end resting on a semi-rough surface, then released. Find the initial acceleration of the center of the rod. (Since the rod is released from rest, ? = 0 at this instant.) The coefficient of kinetic friction between the rod and the surface is 0.21.

a) what is the y-component of the acceleration? (clockwise is positive)

b) what is the x-component of the acceleration?

c) what is the angular acceleration?

Explanation / Answer

c] Angular acceleration = torque/ moment of inertia

= (mg cos theta L/2)/(mL^2/3)

= [ 9.8* cos 40 degree /2] / (0.81/3)

= 13.9 rad/s^2

a] y-component of the acceleration = alpha*L/2* cos 40 degree

= 13.9* cos 40 degree*0.81/2 = 4.31m/s^2

b] Normal N = mg - ma = 0.52*(9.8-4.31) = 2.8548

x component of acceleration a = F/m = uN/m = 0.21*2.8548/0.52 = 1.153 m/s^2

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