A uniform disk with mass 42.2 kg and radius 0.270 m is pivoted at its center abo

Question

A uniform disk with mass 42.2 kg and radius 0.270 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.0 N is applied tangent to the rim of the disk.

Part A What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.100 revolution? Express your answer with the appropriate units.

Part B What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.100 revolution? Express your answer with the appropriate units.

Explanation / Answer

Given that mass m = 42.2 kg

and radius R = 0.27 m

moment of inertia of the disk is I = 0.5*M*R^2 = 0.5*42.2*0.27^2 = 1.54 kg-m^2

Net torque acting is Tnet = R*F*sin(90) = 0.27*34*sin(90)= 9.18 N-m
But Tnet = I*alpha

9.18 = 1.54*alpha

alpha = 9.18/1.54 = 5.96 rad/s^2

Using kinematic equations

w^2 - wo^2 = 2*alpha*theta

w^2 -0^2 = 2*5.96*0.1*2*3.142

w = 2.73 rad/sec

then v = R*w = 0.27*2.73 = 0.74 m/sec is the answer for A)

B) a_net = sqrt(a_rad^2 + a_tan^2)

a_rad = V^2/R = 0.74^2/0.27 = 2.02 m/s^2

a_tan = R*alpha = 0.27*5.96 = 1.6 m/s^2

a_net = sqrt(2.02^2+1.6^2) = 2.57 m/s^2

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