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A disk with mass m = 11.9 kg and radius R = 0.32 m begins at rest and accelerate

ID: 1519780 • Letter: A

Question

A disk with mass m = 11.9 kg and radius R = 0.32 m begins at rest and accelerates uniformly for t = 17.5 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 17.5 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 17.5 seconds?  

m

Explanation / Answer

1)

= / t = 32 rad/s / 17.5s = 1.8286 rad/s^2

2)

= 1/2t² = 1/2 * 1.8286 rad/s^2* (17.5s)^2 = 280 rads

3)

I = 1/2mR^2 = 1/2* 11.9kg *(0.32m)^2 = 0.60928 kg·m^2

4)

Ek = 1/2I^2 = 1/2*0.60928kg·m^2 *(32rad/s)^2 = 312 J

5)

a = R = 1.8286 rad/s^2 *0.32m = 0.585 m/s^2

6)

a = ^2 R = (16rad/s)^2 *0.32m = 81.92 m/s^2

7)

v = R = 32 rad/s * 1/2(0.32m) = 5.12 m/s

8)

s = R = 280 rads * 0.32m = 89.6 m

Please check the calculations...

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