A uniform disk with mass 38.5 kg and radius 0.290 m is pivoted at its center abo

Question

A uniform disk with mass 38.5 kg and radius 0.290 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. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.390 revolution? Express your answer with the appropriate units. What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.390 revolution? Express your answer with the appropriate units.

Explanation / Answer

According to the given problem,

a)First calculater the torque,

Torque = F*r = 25.5N * 0.27m = 7.395 N·m

And we also know that

= I = 1/2mr2,

Calculate the angular accelaration,

7.395 N·m = 0.5* 38.5kg * (0.29m)2 *

= 4.567 rad/s2

Using equation of rotational kinamatics

2 = 02 + 2 = 0 + 2 * 4.567rad/s2 * 0.390rev * 2rads/rev = 22.38 rad/s2

= 4.73 rad/s

v = r = 4.73rad/s * 0.29m = 1.372 m/s

b)Calculate linear values,

tangential at = r = 4.567 rad/s2 * 0.29m = 1.324 m/s2

centripetal ac = 2r = 22.38 * 0.29m = 6.49 m/s2

a = at2 + ac2)

a = 6.624 m/s2

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