A uniform disk with mass 45.0 kg and radius 0.250 m is pivoted at its center abo

Question

A uniform disk with mass 45.0 kg and radius 0.250 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 25.5 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 0.310 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 0.310 revolution?

Explanation / Answer

a)

I = 0.5 m R2 = 0.5 * 45.0 *0.255*0.255 = 1.4630625 Kg.m2

= I = F R

= F R/I = 25.5 * 0.255/1.4630625 = 4.4444 rad/s2

2 = 2

= 0.310 * 2 * = 1.94778 rad

2 = 2 = 2 * 4.4444 * 1.94778 = 17.31342

= 4.16094 rad/s

v = R = 0.255 * 4.16094 = 1.061m/s

b)

a = R 2 = 0.255 * 4.16094* 4.16094 = 4.41 m/s2

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