A cylinder with moment of inertia I 1 = 1 kg.m 2 rotates with angular velocity 1

Question

A cylinder with moment of inertia I1 = 1 kg.m2 rotates with angular velocity 1 = 6 rad/s about a frictionless vertical axle. See “initial” diagram below. A second cylinder, with moment of inertia I2 = 2 kg.m2, initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same angular velocity 2, as shown in “intermediate” below. Next, a third cylinder, with moment of inertia I3 = 3 kg.m2, initially not rotating, drops onto the first two to produce the “final” result shown below, with all three rotating at the same angular velocity 3.

What is the moment of inertia of the “final” system? [3 points]

What are the “intermediate” and “final” angular velocities, 2 and 3? [5

points]

What is the ratio of the total initial to the total final kinetic energy? [5

4. d) Rotational Dynamics. A cylinder with moment of inertia 11 1 kg m2 rotates with angular velocity 6 rad/s about a frictionless vertical axle. See initial" diagram below. A wi second cylinder, with moment of inertia I2 2 kg. initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same angular velocity w2, as shown in "intermediate" below. Next, a third cylinder, with moment of inertia I3 3 kg.m2, tially not rotating, drops onto the first two to produce the "final" result shown below, with all three rotating at the same angular velocity was Final Initial Intermediate (i) What is the moment of inertia of the "final" system /9 points) (ii) What are the "intermediate" and "final" angular velocities, w2 and w3? 5 points (iii) What is the ratio of the total initial to the total final kinetic energy? /5 points/ 10

Explanation / Answer

(1)moment of inertia of final system= i1+ i2+ i3

Itotal=1+2+3= 6 kgm2 .

(2)using the cnservation of angular momentum,

i1 w1 = (i1 + i2) w2

1x6=(1+2)w2

w2= 2 rad/s.

again using the cnservation of angular momentum,

(i1 + i2) w2= (i1+i2+i3)w3

(1+2)* 2 = (1+2+3)w3

6=6*w3

w3= 1 rad/s.

(3)initial kEi= 1/2 i1 w12

KEi= 1/2 (1)(6)2 =18j

final kinetic energy= KEf= 1/2 i3 w32

Kef=1/2 (6)(1)2.=3 j

KEi/KEf= 18/3

KEi/KEf= 6

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