A uniform disk with mass 40.0kg and radius 0.200m is pivoted at its center about

Question

A uniform disk with mass 40.0kg and radius 0.200m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and the a constant force F= 30.0 N is applied tangent to the rim of the disk.

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 .200 revolution?

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 .200 revolution?

Explanation / Answer

Here,

mass of disk , m = 40 Kg

radius , r = 0.20 m

F = 30 N

theta = 0.20 rev = 1.26 rad

a)

let the angular acceleration of disk is a

Using second law of motion

I * a = T

0.5 * m * r^2 * a = F * r

0.5 * 40 * 0.20^2 * a = 30 * 0.20

a = 7.5 rad/s^2

Using third equation of motion

wf^2 - wi^2 = 2 * a * theta

wf^2 = 2 * 7.5 * 1.26

wf = 4.35 rad/s

magnitude of tangential velocity = wf * r

magnitude of tangential velocity = 4.35 * 0.200

magnitude of tangential velocity = 0.85 m/s

the magnitude of tangential velocity is 0.85 m/s

b)

centripetal acceleartion , ac = wf^2 * r

ac = 4.35^2 * 0.20

ac = 3.78 rad/s^2

tangential acceleration, at = a * r

at = 4.35 * 0.20

at = 0.87 rad/s^2

resultant acceleration = sqrt(at^2 + ac^2)

resultant acceleration = 3.88 rad/s^2

the resultant acceleration of a point on the rim of the disk is 3.88 rad/s^2

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