Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ω0, but in opposite directions. The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity.
(a) What is the magnitude of that final angular velocity in terms of ω0? (Use the following as necessary: ω0.)
(b) What is the change in rotational kinetic energy of the system? (Use the following as necessary: K as the initial kinetic energy.)
Explain.
Conservation of rotational momentum applies.
I1*w1+I2*w2= (I1+I2)wf
you know w2=-w1=wo
I1=k*mr1^2
I2=k*mr2^2
Solve for wf, it is just algebra.
To solve this problem, we need to understand the conservation of angular momentum and rotational kinetic energy.
(a) The angular momentum of a rotating object is given by the product of its moment of inertia and its angular velocity. It states that the angular momentum of a system remains constant unless an external torque is applied.
The moment of inertia of a disc is given by I = (1/2) * m * r^2, where m is the mass of the disc and r is its radius. Since the masses of both disks are identical, their moments of inertia are in the ratio:
I1 / I2 = [(1/2) * m * r^2] / [(1/2) * m * (2r)^2] = 1 / 4
When the disks are brought together and the common angular velocity is reached, the angular momentum is conserved:
I1 * ω0 + I2 * (-ω0) = (I1 + I2) * ω_f, where ω_f is the final angular velocity.
Substituting the values of I1 and I2, and solving for ω_f:
(1/2) * m * r^2 * ω0 - (1/2) * m * (2r)^2 * ω0 = (3/2) * m * r^2 * ω_f
ω_f = [(1/2) * r^2 - (1/2) * (2r)^2] / [(3/2) * r^2] * ω0
Simplifying the expression, we get:
ω_f = - ω0 / 2
Therefore, the magnitude of the final angular velocity in terms of ω0 is ω_f = ω0 / 2.
(b) The change in rotational kinetic energy of the system can be determined by subtracting the initial kinetic energy (K) from the final kinetic energy.
The initial kinetic energy of the system is given by:
K_initial = (1/2) * I1 * ω0^2 + (1/2) * I2 * ω0^2
Substituting the values of I1 and I2, we have:
K_initial = (1/2) * m * r^2 * ω0^2 + (1/2) * m * (2r)^2 * ω0^2
= (1/2) * m * r^2 * ω0^2 + 2 * m * r^2 * ω0^2
= (1/2 + 2) * m * r^2 * ω0^2
= (5/2) * m * r^2 * ω0^2
The final kinetic energy can be calculated using the final angular velocity ω_f:
K_final = (1/2) * (I1 + I2) * ω_f^2
Substituting the values of I1 and I2, and the final angular velocity ω_f = - ω0 / 2, we have:
K_final = (1/2) * (m * r^2 + m * (2r)^2) * (- ω0 / 2)^2
= (1/2) * (m * r^2 + 4m * r^2) * (1/4) * ω0^2
= (5/8) * m * r^2 * ω0^2
The change in rotational kinetic energy is then:
ΔK = K_final - K_initial
= [(5/8) * m * r^2 * ω0^2] - [(5/2) * m * r^2 * ω0^2]
= [-(15/8) * m * r^2 * ω0^2]
Therefore, the change in rotational kinetic energy of the system is ΔK = -(15/8) * m * r^2 * ω0^2.