A uniform rod of length L_1 = 2.8 m and mass M = 1.8 kg is supported by a hinge

Question

A uniform rod of length L_1 = 2.8 m and mass M = 1.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L_2 = 2.4 m from the hinge. The particle sticks to the rod on contact. After the collision, theta _max = 45 degree. (a) Find m. _______ kg (b) How much energy is dissipated during the collision? _______ J Assume that there is no friction between the rod and the hinge. Because the net external torque acting on the system at the collision point is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle system swings upward.

Explanation / Answer

Applying energy conservation to find its angular velocity just before it hits the ball,

M g L1 / 2 = ( M L1^2 / 3) w0^2 / 2

24.696 = (4.704) w0^2 / 2

w0 = 3.24 rad/s

Applying angular momentum conservation for the collision,

I1 w0 = (I1 + m L2^2 ) w

4.704 x 3.24 = (4.704 + (m x 2.4^2)) w

15.24 = (4.704 + 5.76m) w ....... (i)

applying energy conservation to find its angular velocity just after ball sticks to the rod,

(( M L1^2 / 3) + (m L2^2)) w^2 /2 = M g (L1/2) (1 - cos45) + m g L2 (1- cos45)

(4.704 + 5.76m) w^2 / 2 = (1.8 x 9.8 x 2.8 / 2) (1 - cos45) + (m x 9.8 x 2.4 ( 1- cos45))

(4.704 + 5.76m) w^2 / 2 = 14.57 + 6.89m .....(ii)

(ii) / (i)^2 => 1 / 2((4.704 + 5.76m)) = 14.57 + 6.89m / 15.24^2

232.26 = 137.07 + 232.67m + 79.37 m^2

79.37 m^2 + 232.67m - 95.19 = 0

m = 0.36 kg ......Ans(a)

(b) initial energy = M g L1 /2 = 24.7 J

final energy = 14.57 + 6.89m = 17.05 J

energy dissipated = Ki - Kf = 7.65 J

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