Question

172

The figure above shows a thin, uniform bar of length D = 1.27 m and mass M = 0.96 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 60°, find the speed of the particle before impact.

Answer 1

moment of inertia of rod at one end = I = Md² / 3

Using conservation of angular momentum

m v x + 0 = mv' x + I W

m v (0.8)d = mv'(0.8)d + (Md² / 3) (v'/x)

0.8 m v = 0.8 m v' + (Md / 3) (v'/x)                         Eq-1

h = height gained = d - d Cos60 = 1.27 - 1.27 Cos60 = 0.635 m

using conservation of energy ::

Potential energy at angle 60 = initial kinetic energy of particle + initial kinetic energy of rod + initial potential energy of rod

mgh + Mgh = (0.5) m v'² + (0.5) I W² + Mg (0.30d)

(mg + Mg ) h = (0.5) (m v'² + (Md² / 3) (v'/x)²) + Mg (0.30d)
(0.30 + 0.96) (9.8) (0.635) = (0.5) (0.30 (1.27)² + (0.96 (1.27)²/3)) (v'/ (0.8 x 1.27 ))²) + 0.96 x 9.8 (0.3 x 1.27)

V' = 2.96 m/s   

using equation 1

0.8 m v = 0.8 m v' + (Md / 3) (v'/x)

0.8 (0.30) v = 0.8 (0.30) (2.96) + (0.96 (1.27) / 3) (2.96/(0.8 x 1.27))    

v = 8.93 m/s