Question

A disk rotates freely on a vertical axis with an angular velocity of 60 rpm. An identical disk rotates above it in the same direction about the same axis, but without touching the lower disk, at 10 rpm. The upper disk then drops onto the lower disk. After a short time, because of friction, they rotate together.The final angular velocity of the disks is ____________.

Answer 1

Answer:

35rpm

Explanation:

$w_{i1}$ = 60rpm

$w_{i2}$ = 10rpm

As both the disks are identical, inertia would be same

therefore,

$I_{1}$=$I_{2}$

The equation of angular momentum will be,

$I w_{1i} + I w_{2i} = I w_{1f} + I w_{2f}$

$I( w_{1i} +w_{21} )= I(w_{1f}+w_{2f} )$ ---> (cancel out I)

Angular speed is same, because of friction, they rotate together.

$w_{1f} = w_{2f} = w_{f}$

$w_{1i} + w_{2i} = w_{f} + w_{f} \\w_{1i} + w_{2i} = 2 w_{f}\\w_{f} = \frac{w_{1i} + w_{2i}}{2}$

$w_{f}$ = (60+10)/2 = 35rpm

The final angular velocity of the disks is  35rpm

Answer 2

Answer: 35 rpm

Explanation:

Given

w1(i) = 60 rpm = 60 * 2π/60 rad/s = 6.284 rad/s

w2(i) = 10 rpm = 10 * 2π/60 rad/s = 1.05 rad/s

The two disk are identical, so, I1 = I2

Also, there are no torques acting on the disk, so, angular momentum will be conserved

L(i)tot = L(f)tot

I1w1(i) + I2w2(i) = I1w1(f) + I2w2(f)

Remember, I1 = I2, so that

Iw1(i) + Iw2(i) = Iw1(f) + Iw2(f)

I [w1(i) + w2(i)] = I [w1(f) + Iw2(f)], I cancel out each other, so

w1(i) + w2(i) = w1(f) + w2(f)

It should be noted also, as a result of the friction, they rotate together, and thus, have the same final speed. Thus

w(f)1 = w(f)2 = w(f), so we have

w1(i) + w2(i) = w(f) + w(f) = 2w(f)

w(f) = [w1(i) + w2(i)] / 2

w(f) = (60 + 10) / 2

w(f) = 70 / 2

w(f) = 35 rpm