Question

A uniform disk with mass 35.2 kg and radius 0.230 mis pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.5 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 0.210 revolution?

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.210 revolution?

Answer 1

here, for the tangential velocity v

Using work energy theorem

let the angular speed is w

0.50 * I * w^2 = Torque * angle

0.50 * 0.50 * 35.2 * 0.23^2 * w^2 = 29.5 * 0.23 * 0.21 * 2pi

solving for w

w = 4.39 rad/s

tangential velocity = 4.39 * 0.23 = 1.01 m/s

the tangential velocity is 1.01 m/s

Now, for the acceleration

for angular acceleration

0.50 * 35.2 * 0.23^2 * alpha = 29.5 * 0.23

alpha = 7.29 rad/s^2

resultant acceleration = sqrt((alpha * r)^2 + (v^2/r)^2)

resultant acceleration = sqrt((7.29 * 0.23)^2 + (1.01^2/.23)^2)

resultant acceleration = 4.74 m/s^2

the resultant acceleration is 4.74 m/s^2