A large stone disk is viewed from above and is initially at rest as seen in the

Q: A large stone disk is viewed from above and is initially at rest as seen in the

moment of inertia = I = (1/2)*M*R^2 torque = I*alfa F*R = I*alfa 40*2.2 = (1/2)*145*2.2^2*alfa alfa = 0.251 rad/s^2 alfa = 0.251 k (b) initial velocity = wi = 0 final angular velocity = wf =wi + alfa*t Wf = 0 + 0.251*62 = 15.562 rad/s wf = 11.562 k (c) Vf = r*wf = 2.2*15.562 = 34.2364 m/s

ID: 1466139 • Letter: A

Question

A large stone disk is viewed from above and is initially at rest as seen in the figure below. The disk has a mass of 145.0 kg and a radius of 2.200 m. A constant force of 40.0 N is applied tangent to the edge of the disk for 62.0 s, causing the disk to spin around the z axis.

(a) Calculate the angular acceleration of the stone, finding both the direction and magnitude. (Express your answer in vector form. Substitute numeric values, do not use variables.)

rad/s2

(b) What is the final angular velocity of the stone? (Express your answer in vector form. Substitute numeric values, do not use variables.)

rad/s

(c) Calculate the translational speed for a point on the edge of the stone after 62.0 s. in
m/s

Explanation / Answer

moment of inertia = I = (1/2)*M*R^2

torque = I*alfa

F*R = I*alfa

40*2.2 = (1/2)*145*2.2^2*alfa

alfa = 0.251 rad/s^2

alfa = 0.251 k

(b)

initial velocity = wi = 0

final angular velocity = wf =wi + alfa*t

Wf = 0 + 0.251*62 = 15.562 rad/s

wf = 11.562 k

(c)

Vf = r*wf = 2.2*15.562 = 34.2364 m/s

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