Question 7, chap 8, sect 5. Assume: When the disk lands on the sur- face it does

Question

Question 7, chap 8, sect 5.

Assume: When the disk lands on the sur- face it does not bounce.

The disk has mass 5 kg and outer radius 60 cm with a radial mass distribution (which may not be uniform) so that its moment of

inertia is 3 m R2 . 5

The disk is rotating at angular speed

7 rad/s around its axis when it touches the

surface, as shown in the figure below. The

disk is carefully lowered onto a horizontal sur-

face and released at time t with zero initial 0

linear velocity along the surface. The coef- ficient of friction between the disk and the surface is 0.07 .

The kinetic friction force between the sur- face and the disk slows down the rotation of the disk and at the same time gives it a hor- izontal acceleration. Eventually, the disk’s linear motion catches up with its rotation, and the disk begins to roll (at time trolling) without slipping on the surface.

The acceleration of gravity is 9.8 m/s2 .

5 kg 60 cm, radius 7 rad/s 3 m r = 0.07 How long t = t the ball to roll without slipping? Answer in units of s. rolling to does it take for

Explanation / Answer

Torque T = I A = F R

Angluar accleration A = FR/I

A = (5* 9.8 * 0.6)/(3/5 * 5 * 0.6^2)

A = 27.22 rad/s^2

now Angular accleration A = dW/.dt

so time dt = dW/dA

dt = 7/(27.22)

dt = 0.257 secs

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