A uniform disk with mass 36.1 kg and radius 0.240 m is pivoted at its center abo

Question

A uniform disk with mass 36.1 kg and radius 0.240 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 33.0 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

torque = I * alpha

33*0.240 = 36.1*(0.240)^2 /2 *alpha

alpha = 7.61772853186 rad/s2

rotation angle = 0.1 * 2 pi rad = 0.2pi rad

0.2pi = 0.5 * 7.61772853186 * t^2

t= 0.40615535727 sec

w = 0.40615535727*7.61772853186 =3.09398125344 rad/s

tangential velocity = w*R = 0.74255550082 m/s

acceleration of point on rim = centripetel

= w^2R = 2.29745279919 rad/s towards center of circle

and

alpha = 7.61772853186 rad/s2 along circular direction

net acceleration = 7.95663731419 rad/s2

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