Two 0550 kg balls are attached to the ends of a thin rod of

Two 0.550 kg balls are attached to the ends of a thin rod of length 61.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (the figure), a 83.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 2.78 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle (deg) will the system rotate before it momentarily stops? Only question (C)plz

Solution

Here,

m1 = 0.550 Kg

L = 0.61 m

m2 = 0.083 Kg

u = 2.78 m/s
a) let the angular speed of the system is w

Using conservation of angular momentum

(m2 * (L/2)^2 + m1 * L^2/12) * w = m2 * (L/2) * v

(0.083 * 0.61^2/4 + 0.550 * 0.61^2/12) * w = 0.083 * 0.61/2 * 2.78

solving for w

w = 2.84 rad/s

b) ratio of kinetic energy = 0.5 * (0.083 * 0.61^2/4 + 0.550 * 0.61^2/12) * 2.84^2/(0.5 * 0.083 * 2.78^2)

ratio of kinetic energy = 0.312

c)

let the angle is theta

Using conservation of energy

0.5 * I * w^2 = m2 * g *L * (1 - cos(theta))

0.5 * (0.083 * 0.61^2/4 + 0.550 * 0.61^2/12) * 2.84^2 = 0.083 * 9.8 * (0.61/2) * (1 - cos(theta))

solving for theta

theta = 53 degree

the angle rotated before stopping for a moment is 53 degree