A uniform disk with mass 38.9 kg and radius 0.230 m is pivoted at its center abo

Question

A uniform disk with mass 38.9 kg and radius 0.230 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 0.290 revolution?

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 0.290 revolution?

Explanation / Answer

torque T = f*r = 33*0.23 = 7.59 Nm

T = I = ½mr²

= 2*7.59/38.9*0.23^2 = 7.38 rad/s^2

² = o² + 2 = 0+2*7.38*0.29 = 4.28

w = 2.07 rad/sec

v = rw

v = 0.23*2.07 = 0.48 m/sec

b) tangential at = r = 7.38*0.23 = 1.69 m/s^2

centripetal ac = ²r = 2.07^2*0.23 = 0.99 m/s^2

mag a = (at² + ac²) = 1.96 m/s^2

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