A putty of mass 0.06 kg is droppe at height h = 1 m above a horizontal rod. two
ID: 1913772 • Letter: A
Question
A putty of mass 0.06 kg is droppe at height h = 1 m above a horizontal rod. two 2 kg balls are attached to the ends of the thin rod, which has a mass of 1 kg and also is 1 m long. The rod can rotate about a horizontal axis.
a) At what speed does teh putty hit the ball on the right end of the thin rod?
b) consider the rod, the balls and the wad as one system, what is the system's initial angular momentum before the hit. L initial? the putty is considered a particle
c)What is the sytem's angular momentum right after the hit, L initial.
d) What is the system's total moment of inertia I final?
e) What is the angular velocity of the rod right after the hit, omega final?
f) What is the kinetic energy of the sytem , KE rod + KE balls + KE putty, right after the hit?
g) IS KE larger or smaller than, or equal to the kinetic energy of the putty wad right before it hits the ball? Prove it.
Explanation / Answer
a) You have dropped the ball from 20.fts 1ft = 0.3 meters initial height = 6 meters Initial energy = Potential energy = m*g*20 The energy at som height = m*g*h + 1/2*m*v^2 Now, using conservation of energy : m*g*6 = m*g*h + 1/2*m*v^2 and if they should be half of its energy kinetic and halg potential 2m*g*h = m*g*6 h = 3 meters = 10ft answer for a) = 10 fts = 3 meters b) Using conservation of energy : Initial energy = m*g*6 final energy = 1/2*m*v^2 1/2*m*v^2 = m*g*6 v = 10.8 m/s Hope that helps