A uniform disk with mass 44.6kg and radius 0.270m is pivoted at its center about

Question

A uniform disk with mass 44.6kg and radius 0.270m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest and then a constant force 26.5N is applied tangent to the rim of the disk Part A What is the magnitude y of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 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 0.400 revolution?

Explanation / Answer

Note that

Torque = F R = I * alpha

--> alpha = F R / I

As

I = M R^2 / 2

for a disk, then

alpha = 4.4013 rad/s^2

As

wf^2 = wi^2 + 2*alpha*theta

As wi = 0, theta = 0.400 rev = 2.51 rad, solving for wf,

wf = 4.70 rad/s

As v = r w,

vf = 1.27 m/s   [ANSWER, PART A]

*********************

The tangential acceleration is

atan = alpha*r = 1.188 m/s^2

The radial acceleration is

arad = v^2/r = 5.974 m/s^2

Thus, getting the resultant by Pythagorean theorem,

a = 6.09 m/s^2   [ANSWER, PART B]

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