Question

A uniform rod of length 20 cm is freely pivoted at its centre. A gum of mass 0.2 kg moving at a speed of 10 m s-¹ strikes and sticks to one end of the rod. The rod rotates horizontally. The moment of inertia of the rod and gun about the pivot is 0.15 kg m²
a)Calculate the angular momentum of the gum and find the final angular velocity of the rod.
b)If the length of the rod is doubled and the moment of inertia of the rod and the gum is 0.20 kg m²,calculate the new angular velocity of the system.

Answer 1

Answer:

Explanation:

It is given here,

mass of gum, m = 0.2 kg

Speed of gum, v = 10 m/s

length of a rod, l = 20 cm = 0.2 m

The moment of inertia of the rod and gun about the pivot, i = 0.15 kg m²

a) The angular momentum of the gum:

L = mv (1/2*l)  

= 0.2 * 10 * 0.5 * 0.2

= 0.2 kg m² / s

=> Final angular velocity of the rod :

i * ω= l

ω = l / i

= 0.2 / 0.15

= 1.3 rad / s

b)  The new angular velocity of the system when the length of the rod is doubled and the moment of inertia of the rod and the gum is 0.20 kg m² :

ω' = 2* l/ i

= 2 * 0.2 / 0.2

= 2 rad / s