A circular disk of mass 0.2 kg and radius 24 cm, initially not rotating, slips down a thin spindle onto a turntable (disk) of mass 1.7 kg and the same radius, rotating freely at 3.4 rad/s.
a) Find the new angular velocity of the combination;
rad/s
b) The change in the kinetic energy?
c) If the motor is switched on after the disk has landed, what is the constant torque needed to regain the original speed in 2.6 s?
To answer these questions, we can use the principles of conservation of angular momentum and conservation of mechanical energy.
a) To find the new angular velocity of the combination, we can conserve angular momentum. The initial angular momentum of the disk before it slips onto the turntable is zero since it is not rotating. The final angular momentum of the combined system (disk + turntable) is given by the sum of the initial angular momentum of the turntable and the angular momentum acquired by the disk after slipping onto the turntable.
The initial angular momentum of the turntable is given by:
L_initial = I_turntable * ω_initial
where I_turntable is the moment of inertia of the turntable and ω_initial is its initial angular velocity.
The angular momentum acquired by the disk after slipping onto the turntable is given by:
L_acquired = I_disk * ω_combined
where I_disk is the moment of inertia of the disk and ω_combined is the final angular velocity of the combined system.
Since we assume that the disk and turntable are mounted on the same axis, the total moment of inertia of the combined system is:
I_combined = I_disk + I_turntable
Using the conservation of angular momentum, we have:
L_initial = L_acquired
I_turntable * ω_initial = (I_disk + I_turntable) * ω_combined
Substituting the given values, we have:
(0.5 * m_turntable * r^2) * ω_initial = ((0.5 * m_disk * r^2) + (0.5 * m_turntable * r^2)) * ω_combined
Simplifying the equation, we get:
m_turntable * ω_initial = (m_disk + m_turntable) * ω_combined
Now we can plug in the values:
ω_combined = (m_turntable * ω_initial) / (m_disk + m_turntable)
= (1.7 kg * 3.4 rad/s) / (0.2 kg + 1.7 kg)
Calculating this, we get:
ω_combined = 0.527 rad/s
b) To find the change in kinetic energy, we can use the conservation of mechanical energy. The initial kinetic energy is zero since the disk is not rotating, and the final kinetic energy is given by the sum of the initial kinetic energy of the turntable and the kinetic energy acquired by the disk after slipping onto the turntable.
The initial kinetic energy of the turntable is:
K_initial = 0.5 * I_turntable * ω_initial^2
The kinetic energy acquired by the disk after slipping onto the turntable is:
K_acquired = 0.5 * I_disk * ω_combined^2
The total change in kinetic energy is:
ΔK = K_acquired - K_initial = 0.5 * (I_disk * ω_combined^2 - I_turntable * ω_initial^2)
Substituting the given values, we have:
ΔK = 0.5 * ((0.5 * m_disk * r^2) * ω_combined^2 - (0.5 * m_turntable * r^2) * ω_initial^2)
Calculating this, we get:
ΔK = -0.670 J (negative sign indicates a decrease in kinetic energy)
c) If the motor is switched on after the disk has landed, the torque required to regain the original speed can be calculated using the equation:
τ = ΔL / Δt
Where τ is the torque, ΔL is the change in angular momentum, and Δt is the time interval for which the torque is applied.
The change in angular momentum is given by:
ΔL = I_combined * (ω_combined - ω_initial)
Substituting the given values, we have:
ΔL = (I_disk + I_turntable) * (ω_combined - ω_initial)
The time interval is given as Δt = 2.6 s.
Calculating this, we get:
τ = ΔL / Δt = [(0.5 * m_disk * r^2) + (0.5 * m_turntable * r^2)] * (ω_combined - ω_initial) / Δt
Substituting the values, we get:
τ = [(0.5 * 0.2 kg * (0.24 m)^2) + (0.5 * 1.7 kg * (0.24 m)^2)] * (0.527 rad/s - 0) / 2.6 s