A disk with mass m = 10.7 kg and radius R = 0.36 m begins at rest and accelerate

Question

A disk with mass m = 10.7 kg and radius R = 0.36 m begins at rest and accelerates uniformly for t = 17.8 s, to a final angular speed of = 30 rad/s.

1. What is the angular acceleration of the disk?

2. What is the angular displacement over the 17.8 s?

3. What is the moment of inertia of the disk?

4. What is the change in rotational energy of the disk?

5. What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

6. What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

7. What is the final speed of a point on the disk half-way between the center of the disk and the rim?

8. What is the total distance a point on the rim of the disk travels during the 17.8 seconds?

Explanation / Answer

1.)
Initial angular speed, o = 0
Final angular speed , f = 30 rad/s
t = 17.8 s

= - t
= 30/17.8
= 1.685 rad/s^2

Angular acceleration of the disk, = 1.685 rad/s^2

2.)
² = ² + 2
30^2 = 0 + 2*1.685*
= 267.06 rad
Angular displacement, = 267.06 rad

3.)
Moment of inertia of disk, I = 1/2 * m*r^2
I = 1/2 *10.7 * 0.36^2
I =  0.693 Kg m^2

4.)
Change in Rotational Energy of the disk = Rotational K.Efin - Rotational K.Ein
Change in Rotational Energy of the disk = 1/2 * I*wf^2 - 1/2 * I*wo^2
Change in Rotational Energy of the disk = 1/2 * 0.693*30^2 - 0
Change in Rotational Energy of the disk = 311.9 J

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