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
Answer:
35rpm
Explanation:
[tex]w_{i1}[/tex] = 60rpm
[tex]w_{i2}[/tex] = 10rpm
As both the disks are identical, inertia would be same
therefore,
[tex]I_{1}[/tex]=[tex]I_{2}[/tex]
The equation of angular momentum will be,
[tex]I w_{1i} + I w_{2i} = I w_{1f} + I w_{2f}[/tex]
[tex]I( w_{1i} +w_{21} )= I(w_{1f}+w_{2f} )[/tex] ---> (cancel out I)
Angular speed is same, because of friction, they rotate together.
[tex]w_{1f} = w_{2f} = w_{f}[/tex]
[tex]w_{1i} + w_{2i} = w_{f} + w_{f} \\w_{1i} + w_{2i} = 2 w_{f}\\w_{f} = \frac{w_{1i} + w_{2i}}{2}[/tex]
[tex]w_{f}[/tex] = (60+10)/2 = 35rpm
The final angular velocity of the disks is 35rpm
