A uniform disk with mass 40.8 kg and radius 0.300 m is pivoted at its center abo

Question

A uniform disk with mass 40.8 kg and radius 0.300 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 26.0 N is applied tangent to the rim of the disk.

What is the magnitude "v" of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolution?

What is the magnitude "a" of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.320 revolution?

Explanation / Answer

= I

0.3 X 26 = (40.8 x 0.32 /2)

= 4.25 rad/s2

F is constant , so a & both will also be constant.

a = r = 4.25 X 0.3 = 1.27 m/s2 ..Ans

= 0.320 rev. = 2.01 rad

2 = 2

   = 2 X 4.25 X 2.01= 17.09

= 4.13 rad/s

v = r = 4.13 x 0.3 = 1.24 m/s .........Ans

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