Exercise 10.10 Part A What is the magnitude of the tangential velocity of a poin

Question

Exercise 10.10 Part A What is the magnitude of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.120 revolution? A uniform disk with mass 44.2 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 28.5 N is applied tangent to the rim of the disk Express your answer with the appropriate units. 1.944 m Submit My Answers Give Up Incorrect; One attempt remaining ; One attempt remaining: Try Again 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 0.120 revolution? Express your answer with the appropriate units. = 2.33 Submit My Answers Give Up Correct

Explanation / Answer

here,

mass , m = 44.2 kg

radius , r = 0.25 m

constant force , F = 28.5 N

A)

let the accelration be alpha

I * alpha = F * r

0.5 * 44.2 * 0.25^2 * alpha = 28.5 * 0.25

alpha = 5.16 rad/s^2

angle turned , theta = 0.12 rad

using third equation of motion

w = sqrt( 2 * alpha * theta)

w = 1.11 rad/s

the magnitude of the speed of rim , v = r * w

v = 1.11 * 0.25 = 0.28 m/s

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