A disk with mass m = 8.9 kg and radius R = 0.37 m begins at rest and accelerates

Question

A disk with mass m = 8.9 kg and radius R = 0.37 m begins at rest and accelerates uniformly for t = 16.7 s, to a final angular speed of = 32 rad/s.

1)

What is the angular acceleration of the disk?

rad/s2  

2)

What is the angular displacement over the 16.7 s?

rad  

3)

What is the moment of inertia of the disk?

kg-m2  

4)

What is the change in rotational energy of the disk?

J  

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?

m/s2  

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?

m/s2

7)

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

m/s  

8)

What is the total distance a point on the rim of the disk travels during the 16.7 seconds?

Explanation / Answer

The problem can be solved as follows -

(1) = / t = 32 rad/s / 16.7 s = 1.92 rad/s²

(2) = ½t² = ½ * 1.92 rad/s² * (16.7 s)² = 267.7 rads

(3) I = ½mr² = ½ * 8.9kg * (0.37m)² = 0.61 kg·m²

(4) Ek = ½I² = ½ * 0.61kg·m² * (32rad/s)² = 312 J

(5) a = r = 1.92rad/s² * 0.37m = 0.71 m/s²

(6) a = ² r = (16rad/s)² * 0.37m = 94.7 m/s²

(7) v = r = 32 rad/s * ½(0.37m) = 5.92 m/s

(8) s = r = 267.7 rads * 0.37m = 99 m

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