The figure below shows a thin, uniform bar of length D = 1.15 m and mass M = 0.7

Question

The figure below shows a thin, uniform bar of length D = 1.15 m and mass M = 0.71 kg pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m = 0.30 kg at a point x = 0.80d below the pivot. Assume that the particle sticks to the rod. If the maximum angle between the rod and the vertical following the collision is 60degree, find the speed of the particle before impact. 5.22 m/s Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. You can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing. Draw a sketch showing the bar and particle just before the collision, and when the bar is at its maximum angle.

Explanation / Answer

D = 1.15 m ; M = 0.71 kg ; m = 0.3 kg ; x = 0.8 d

The total moment of inertis of the given system will be:

I = 1/3 M d^2 + m x^2

I = 0.71 x 1.15^2/3 + 0.3 x 0.8^2 = 0.505 kg-m^2

Initial angular momentum of the system: Li = m v x

Final angular momentum of the system: Lf = I w
  
from conservation of angular momentum,

Li = Lf

m v x = I w

v = I w/ m x (1)

from energy conservation:

Ei = Ef

1/2 I w^2 = Mgd/2 (1 - cos(theta)) + mgx(1 - cos(theta))

0.5 x 0.505 x w^2 = 0.71 x 9.8 x 1.15 x 0.5 (1 - cos60) + 0.3 x 9.8 x 0.8 (1 - cos60)

0.253 w^2 = 2 + 1.18

w = 3.55 rad/s

putting this result in (1)

v = 0.505 x 3.55/0.3 x 0.8 = 7.47 m/s

Hence, v = 7.47 m/s

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