A child of mass 45 kg stands beside a circular platform of mass 85 kg and radius 1.7m spinning at 4 rad/s. Treat the platform as a disk. The child steps on the rim.
a) What is the new angular speed; (rad/s)
b) She then walks to the center and stay there. What is the angular velocity of the platform then? (rad/s)
c) What is the change in kinetic energy when she walks from the rim to the center of the platform? (J)
Initial angular momentum
= (1/2) m R^2 omega
= (1/2)85 * 1.7^2 * 4
final angular momentum with kid on rim
= [(1/2) 85*1.7^2 + 45*1.7^2]*new omega
there is no torque, so no change in angular momentum
so
solve for new omega (part a) (turns slower)
then
kid walks to center
still no change in angular momentum but the moment of inertia reverts to the original so the omega goes back up to the original
energy kid on rim = (1/2) [(1/2) 85*1.7^2 + 45*1.7^2]* (new omega)^2
energy kid at center = (1/2)[(1/2) 85*1.7^2] *(original omega)^2
To solve this problem, we can use the principle of conservation of angular momentum and the principle of conservation of energy.
a) To find the new angular speed when the child steps on the rim of the platform, we can use the conservation of angular momentum equation:
Initial angular momentum = Final angular momentum
The initial angular momentum of the system is given by the circular platform spinning at 4 rad/s:
I1 * ω1 = (1/2) * M * R^2 * ω1,
where I1 is the moment of inertia of the platform, M is the mass of the platform, R is the radius of the platform, and ω1 is the initial angular velocity.
The final angular momentum is given by the combined system of the platform and the child standing on the rim. The moment of inertia of the combined system is:
I2 = (1/2) * M * R^2 + M_child * R^2,
where M_child is the mass of the child.
Let ω2 be the new angular velocity.
Therefore, the equation becomes:
(1/2) * M * R^2 * ω1 = [(1/2) * M * R^2 + M_child * R^2] * ω2,
Substituting the given values: M = 85 kg, R = 1.7 m, ω1 = 4 rad/s, and M_child = 45 kg,
(1/2) * 85 * (1.7^2) * 4 = [(1/2) * 85 * (1.7^2) + 45 * (1.7^2)] * ω2,
1129 = (878.625 + 1291.25) * ω2,
1129 = 2169.875 * ω2,
ω2 = 1129 / 2169.875,
ω2 ≈ 0.520 rad/s.
Therefore, the new angular speed when the child steps on the rim is approximately 0.520 rad/s.
b) When the child walks to the center and stays there, the moment of inertia of the combined system is now:
I3 = M * R^2 + M_child * (R/2)^2,
where R/2 is the new radius since the child is now at the center of the platform.
Let ω3 be the new angular velocity.
Using the conservation of angular momentum, the equation becomes:
[(1/2) * M * R^2 + M_child * R^2] * ω2 = [M * R^2 + M_child * (R/2)^2] * ω3,
Simplifying the equation:
[(1/2) * 85 * (1.7^2) + 45 * (1.7^2)] * 0.520 = [85 * (1.7^2) + 45 * (1.7/2)^2] * ω3,
1291.25 * 0.520 = [2204.75 + 265.42625] * ω3,
670.075 = 2470.17625 * ω3,
ω3 = 670.075 / 2470.17625,
ω3 ≈ 0.271 rad/s.
Therefore, the angular velocity of the platform when the child walks to the center and stays there is approximately 0.271 rad/s.
c) The change in kinetic energy is given by:
Change in kinetic energy = Final kinetic energy - Initial kinetic energy,
The initial kinetic energy is:
Initial kinetic energy = (1/2) * I1 * ω1^2,
Substituting the given values: I1 = (1/2) * 85 * (1.7^2), and ω1 = 4 rad/s,
Initial kinetic energy = (1/2) * [(1/2) * 85 * (1.7^2)] * (4^2),
Initial kinetic energy ≈ 483.2 J.
The final kinetic energy is:
Final kinetic energy = (1/2) * I3 * ω3^2,
Substituting the given values: I3 = 85 * (1.7^2) + 45 * (1.7/2)^2, and ω3 ≈ 0.271 rad/s,
Final kinetic energy = (1/2) * [85 * (1.7^2) + 45 * (1.7/2)^2] * (0.271^2),
Final kinetic energy ≈ 29.2 J.
Therefore, the change in kinetic energy when the child walks from the rim to the center of the platform is approximately 29.2 J - 483.2 J = -454 J (since the final kinetic energy is less than the initial kinetic energy).
To answer these questions, we need to apply the principles of conservation of angular momentum and conservation of kinetic energy.
a) To find the new angular speed when the child steps on the rim, we can use the principle of conservation of angular momentum. The initial angular momentum of the system, consisting of the child and the platform, is given by:
L_initial = I_initial * ω_initial
where L_initial is the total angular momentum, I_initial is the moment of inertia of the system, and ω_initial is the initial angular speed.
The moment of inertia of a solid disk is given by:
I = (1/2) * m * r^2
where m is the mass of the platform (85 kg), and r is the radius of the platform (1.7 m).
Substituting the values, we have:
I_initial = (1/2) * 85 * (1.7)^2 = 144.85 kg·m^2
Also, the initial angular speed is given as ω_initial = 4 rad/s.
The final angular momentum of the system when the child steps on the rim is:
L_final = I_final * ω_final
where L_final is the final angular momentum, I_final is the moment of inertia of the system with the child on the rim, and ω_final is the new unknown angular speed.
Since the system is still rotating after the child steps on the rim, the total angular momentum is conserved, meaning L_initial = L_final. Therefore, we can set up the equation:
I_initial * ω_initial = I_final * ω_final
Solving for ω_final, we have:
ω_final = (I_initial * ω_initial) / I_final
To calculate I_final, we need to add the moment of inertia of the child to the initial moment of inertia of the system:
I_final = I_initial + (1/2) * m_child * r^2
where m_child is the mass of the child (45 kg).
Substituting the values, we can calculate ω_final to find the new angular speed.
b) After the child walks to the center of the platform and stays there, the mass distribution of the system changes. Now, the child's mass is concentrated at the center, and the platform's moment of inertia becomes that of a circular disk. The moment of inertia of a circular disk is given by:
I_disk = (1/2) * m * r^2
where m is the mass of the platform (85 kg) and r is the new radius (0 m for a central point).
The angular velocity of the platform when the child is at the center can be calculated using the equation:
L_final = I_final * ω_final
where L_final is the final angular momentum, I_final is the moment of inertia of the platform alone, and ω_final is the final angular velocity.
c) The change in kinetic energy when the child walks from the rim to the center can be calculated using the conservation of kinetic energy principle. The initial kinetic energy is given by:
KE_initial = (1/2) * I_initial * ω_initial^2
where KE_initial is the initial kinetic energy, I_initial is the initial moment of inertia, and ω_initial is the initial angular speed.
The final kinetic energy is given by:
KE_final = (1/2) * I_final * ω_final^2
where KE_final is the final kinetic energy, I_final is the final moment of inertia, and ω_final is the final angular speed.
The change in kinetic energy is then given by:
ΔKE = KE_final - KE_initial
Substituting the values and calculating will give you the change in kinetic energy.
To summarize:
a) Use the principle of conservation of angular momentum to find the new angular speed when the child steps on the rim.
b) Use the principle of conservation of angular momentum again to find the angular velocity when the child is at the center.
c) Use the principle of conservation of kinetic energy to find the change in kinetic energy when the child walks from the rim to the center of the platform.