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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