. A disk slides toward a motionless stick on a frictionless surface.
b. The disk hits the stick at one end and adheres to it, and they rotate together, pivoting around the nail.
Suppose the disk in the picture above has a mass of 50.0 g and an initial velocity of 30 * m/s when it strikes the stick that is 1.20 m long and 2.00 kg at a point 1.00 m from the pivot. What is the angular velocity (in rad s ) of the two after the collision?
To solve this problem, we can use the principle of conservation of angular momentum.
The initial angular momentum of the system is given by:
L_initial = m_initial * v_initial * r_initial
where:
m_initial = mass of the disk = 50.0 g = 0.05 kg
v_initial = initial velocity of the disk = 30 m/s
r_initial = initial distance of the disk from the pivot = 1.00 m
L_initial = 0.05 kg * 30 m/s * 1.00 m
L_initial = 1.5 kg m^2/s
Since the stick was initially at rest, the total initial angular momentum of the system is equal to L_initial.
After the collision, the disk sticks to the stick and they rotate together around the nail. The final moment of inertia of the system is the sum of the moments of inertia of the disk and the stick:
I_final = m_disk * r_disk^2 + m_stick * r_stick^2
where:
m_disk = mass of the disk = 0.05 kg
r_disk = distance of the disk from the pivot = 1.00 m
m_stick = mass of the stick = 2.00 kg
r_stick = distance of the center of mass of the stick from the pivot = 1.00 m
I_final = 0.05 kg * (1.00 m)^2 + 2.00 kg * (1.00 m)^2
I_final = 2.05 kg m^2
The final angular momentum of the system is equal to the initial angular momentum, so:
L_final = I_final * ω_final
where:
ω_final = angular velocity of the system after the collision
Substitute in the values:
1.5 kg m^2/s = 2.05 kg m^2 * ω_final
ω_final = 1.5 kg m^2/s / 2.05 kg m^2
ω_final = 0.7317 rad/s
Therefore, the angular velocity of the system after the collision is 0.7317 rad/s.