The figures above show a disk with radius, a = 0.20 m, and mass, M = 0.80 kg, re

Question

The figures above show a disk with radius, a = 0.20 m, and mass, M = 0.80 kg, resting on a frictionless table. One particle with mass, m1 = M/4, with velocity, v = 4 m/s, slides along the stable, and collides with the disk at the point shown. A second particle with mass, m2, moving with velocity v2 = v4 collides with the disk at the point shown. The two masses collide with the disk at the same time, and after the collision, and point particles stick on the disk. We will consider the two situations shown in Figs. A and B.

For this part, the situation is as shown in Fig. A with the disk is free to move on the table.

a) What is m2 if the center of mass does not move before or after the collision?  

For parts b-d, assume that the situation is as shown in Fig. B with the same particles colliding with the same disk, but now there is a pin in the center of the disk so that the disk must rotate about its center. Take m2 to be the value you obtained in part a.

b) What is the initial angular momentum, L0, of the system?

c) What is the rotational inertial, I of the disk and masses after the collision?

Explanation / Answer

Vcm = ( (m1*v1) + (m2*v2))/(m1+m2)

as the center of mass is not moving

Vcm = 0

0 = (M/4*4)-(m2*4)/(M/4 + m2)

M/4*4 = 2*m2

m2 = M/2 <<-----answer

+++++++++++++++++++

b)

Lo = m1*v1*a + m2*v2*a*cos60

Lo = ((0.8/4*4*0.2) + ((0.8/2)*4*0.2*cos60)

Lo = 0.32 kg m^2 /s

c)

I = (1/2)*M*a^2 + m1*a^2 + m2*a^2

I = (1/2)*0.8*0.2^2 + (0.8/4)*0.2^2 + (0.8/2)*0.2^2

I = (0.4+0.2+0.4)*0.2^2 = 0.04 kg m^2

d)

Lf = I*w

from momentum conservation

Lf = Lo

0.04*w = 0.32

w = 8 rad/s

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