A uniform disk with mass 39.1 and radius 0.200 is pivoted at its center about a

Question

A uniform disk with mass 39.1 and radius 0.200 is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 30.5 is applied tangent to the rim of the disk.What is the magnitude of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.130 revolution?
What is the magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.130 revolution?

Explanation / Answer

= 30.5*.2 = I = (.5*39.1*.2^2)*

so, =7.8

so, = sqrt(2*7.8*.13*2*) = 3.57

so, tangential velocity = 3.57*.2 = 0.714 m/s <-----answer

so, centripetal acceleration = .714^/.2 = 2.55 m/s2

tangential acc = 7.8*.2 = 1.56 m/s2

so, net acceleration = sqrt(2.55^2 + 1.56^2) = 2.99 m/s^2 <----answer

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