All the answers are given. I just need step by step solutions please. 1) Three p
ID: 2243118 • Letter: A
Question
All the answers are given. I just need step by step solutions please.
1) Three particles each of mass m are fastened to each other and to a rotation axis at O by three massless strings, each of length L. At t=0, the three masses have coordinates (L,0), (2L,0) and (3L,0). The combination rotates about the z axis with angular velocity v in such a way that the particles remain in a straight line, that is, at a certain time later their coordinates are (0,L), (0,2L) and (0,3L). In terms of m, L and v, and relative to the origin, what are:
a) the moment of inertia of the combination? (14 mL^2)
b) the angular momentum of the middle particle? (4 mL^2v)
c) the total angular momentum of the three particles? (14 mL^2v)
2) A uniform thin rod of length 0.50 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its centre. The rod is at rest when a 3.00 g bullet traveling in the horizontal plane of the rod is fired into end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of 60 degrees with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact? (1300 m/s)
(Hint: find the moment of inertia for the rod plus bullet, then use conservation of angular momentum. Don't forget to find the component of the bullet's velocity perpendicular to the rod.)
Explanation / Answer
2 The axis of rotation is in the middle of the rod, r = 0.25 m from either end.
By Eq. , the initial angular momentum of the system (which is just that of the bullet, before impact) is rmv sin? where
m = 0.003 kg and ? = 60?. Relative to the axis, this is counterclockwise and thus (by the common
convention) positive. After the collision, the moment of inertia of the system is I = Irod + mr2 where
Irod = ML2/12 by Table 11-2(e), with M = 4.0 kg and L = 0.5 m. Angular momentum conservation
leads to
rmv sin? = ( 1 /12 ML2 + mr2 ) ? .
Thus, with ? = 10 rad/s, we obtain
v = 1/12(4.0)(0.5)^2 + (0.003)(0.25)^2)(10) / (0.25)(0.003) sin 60?
= 1.3