A child of mass 40 kg stands beside a circular platform of mass 90 kg and radius 1.6 m spinning at 2.8 rad/s. Treat the platform as a disk. The child steps on the rim.
a) What is the new angular speed;
b) She then walks to the center and stay there. What is the angular velocity of the platform then?
c) What is the change in kinetic energy when she walks from the rim to the center of the platform?
mindf*ck
To calculate the new angular speed of the platform after the child steps on the rim, we can use the law of conservation of angular momentum.
Angular momentum is defined as the product of moment of inertia and angular velocity. The equation is L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.
a) Initially, we can calculate the angular momentum of the platform-child system. Since the child is not initially on the platform, the moment of inertia will only be for the platform.
The moment of inertia of a disk is given by the equation I = (1/2) * m * r^2, where m is the mass of the platform and r is the radius of the platform.
I = (1/2) * 90 kg * (1.6 m)^2 = 115.2 kg·m^2
The initial angular momentum (L_initial) is calculated by multiplying the moment of inertia by the initial angular velocity (ω_initial) of the platform.
L_initial = I * ω_initial = 115.2 kg·m^2 * 2.8 rad/s = 322.56 kg·m^2/s.
When the child steps on the rim, no external torques are present, so the total angular momentum should be conserved.
The final angular momentum (L_final) will be the sum of the angular momentum of the platform and the child.
L_final = (I * ω) + (m_child * r * ω), where m_child is the mass of the child and r is the radius of the platform.
Since the child is standing on the rim of the platform, the radius (r) remains the same.
Given that the mass of the child (m_child) is 40 kg and the radius (r) is 1.6 m, we can calculate the new angular velocity (ω_final) using the equation:
L_final = (115.2 kg·m^2 * 2.8 rad/s) + (40 kg * 1.6 m * ω_final)
322.56 kg·m^2/s = 322.56 kg·m^2/s + (40 kg * 1.6 m * ω_final)
Now, we can solve for ω_final:
0 = (40 kg * 1.6 m * ω_final)
This implies that ω_final = 0 rad/s.
Therefore, the new angular speed of the platform after the child steps on the rim is 0 rad/s.
b) When the child walks to the center and stays there, the moment of inertia of the system changes. The moment of inertia of the platform-child system can be calculated using the equation I = (1/2) * (90 kg + 40 kg) * (1.6 m)^2, where the mass of the platform and the child are considered.
I = (1/2) * 130 kg * (1.6 m)^2 = 166.4 kg·m^2
Since the child is at the center, the radius (r) is now 0.
Using the conservation of angular momentum, we know that L_initial = L_final.
So, I_initial * ω_initial = I_final * ω_final.
(115.2 kg·m^2 * 2.8 rad/s) = (166.4 kg·m^2 * ω_final).
Now, we can solve for ω_final:
2.8 rad/s = ω_final.
Therefore, the angular velocity of the platform when the child walks to the center and stays there is 2.8 rad/s.
c) To calculate the change in kinetic energy, we need to find the initial kinetic energy (KE_initial) and the final kinetic energy (KE_final).
The initial kinetic energy (KE_initial) is given by the equation KE_initial = (1/2) * I_initial * (ω_initial)^2.
KE_initial = (1/2) * 115.2 kg·m^2 * (2.8 rad/s)^2 = 227.58 J.
The final kinetic energy (KE_final) is given by the equation KE_final = (1/2) * I_final * (ω_final)^2.
KE_final = (1/2) * 166.4 kg·m^2 * (2.8 rad/s)^2 = 734.21 J.
The change in kinetic energy (ΔKE) is calculated by subtracting the initial kinetic energy from the final kinetic energy.
ΔKE = KE_final - KE_initial = 734.21 J - 227.58 J = 506.63 J.
Therefore, the change in kinetic energy when the child walks from the rim to the center of the platform is 506.63 J.