Answer:
The loss in rotational kinetic energy due to the collision is 36.585 J.
Explanation:
Given;
mass of the disk, m₁ = 1.64 kg
radius of the disk, r = 0. 61 m
angular velocity of the disk, ω₁ = 17.6 rad/s
mass of the rod, m₂ = 1.51 kg
length of the rod, L = 1.79 m
angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)
let the counter-clockwise be the positive direction
let the clock-wise be the negative direction
The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;
m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)
where;
ωf is the common final angular velocity
1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)
21.1328 = ωf(3.15)
ωf = 21.1328 / 3.15
ωf = 6.709 rad/s
The moment of inertia of the disk is calculated as follows;
The moment of inertia of the rod about its center is calculated as follows;
The initial rotational kinetic energy of the disk and rod;
The final rotational kinetic energy of the disk-rod system is calculated as follows;
The loss in rotational kinetic energy due to the collision is calculated as follows;
Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.