A cylinder with moment of inertia I 1 rotates with angular velocity ? 0 about a
ID: 1286683 • Letter: A
Question
A cylinder with moment of inertia I1 rotates with angular velocity ?0 about a frictionless vertical axle. A second cylinder, with moment of inertia I2, initially not rotating, drops onto the first cylinder (Fig. P8.55). Because the surfaces are rough, the two eventually reach the same angular velocity, ?.
(a) Calculate ?. (Use I1 for I1, I2 for I2, and w0 for ?0 in your equation.)
(b) Show that energy is lost in this situation (Do this on paper. Your instructor may ask you to turn in this work.), and calculate the ratio of the final to the initial kinetic energy. (Use I1 for I1, I2 for I2, w for ?, and w0 for ?0 in your equation.)
Explanation / Answer
Well, for part a) you need to invoke the conservation of angular momentum.
L_total = L_1 + L_2
L_total = I_1*w1 + I_2*w2
In your case w1 is given as w0, and w2 is 0 (second disk is not spinning), so we get
I_tot*w_tot = I_1*w0
and I_tot is just I_1+I_2, which gives:
w_tot = (w_0 * I_1)/(I_1+I_2)
b)
you just need to compare the energies before and after.
before:
E1 = 1/2 * I1 * w0^2
after:
Etot = 1/2 * I_tot * wtot^2
plug in the values from part a and you are there
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