A uniform disk with mass 43.1 kg and radius 0.230 m is pivoted at its center abo

Question

A uniform disk with mass 43.1 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 35.0 N is applied tangent to the rim of the disk. 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 0.150 revolution? 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 0.150 revolution?

Explanation / Answer

A) using the relation, W^2 = u^2 + 2as , we get W = sqrt(2*(0.23*35/(0.5*43.1*0.23^2))*0.15*2*3.14) = 3.65 rad/s so, tangential velocity, v = 3.65*0.23 = 0.84 m/s

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