A child of mass 40 kg stands beside a circular platform of mass 115 kg and radius 1.6 m spinning at 3.8 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?
a) The new angular speed is 3.2 rad/s.
b) The angular velocity of the platform is 2.8 rad/s.
c) The change in kinetic energy when she walks from the rim to the center of the platform is -2.4 J.
To solve this problem, we can apply the conservation of angular momentum and the conservation of kinetic energy.
a) To find the new angular speed when the child steps on the rim, we can use the conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum.
The initial angular momentum is given by:
(Linitial) = Iplatform * winitial
where Iplatform is the moment of inertia of the platform and winitial is the initial angular speed.
The final angular momentum is given by:
(Lfinal) = (Iplatform + Ichilddisk) * wfinal
where Ichilddisk is the moment of inertia of the child when treated as a disk.
Since there is no external torque acting on the system, the initial and final angular momentum are equal, so we can equate them:
Iplatform * winitial = (Iplatform + Ichilddisk) * wfinal
Substituting the given values:
115 kg * (1.6 m)^2 * 3.8 rad/s = (115 kg * (1.6 m)^2 + 40 kg * (0.8 m)^2) * wfinal
Simplifying:
928.64 kg*m^2/s = (296.96 kg*m^2 + 12.8 kg*m^2) * wfinal
wfinal = 928.64 kg*m^2/s / (296.96 kg*m^2 + 12.8 kg*m^2)
wfinal = 928.64 kg*m^2/s / 309.76 kg*m^2
wfinal ≈ 2.999 rad/s
Therefore, the new angular speed when the child steps on the rim is approximately 2.999 rad/s.
b) When the child walks to the center and stays there, her contribution to the moment of inertia becomes negligible compared to the entire platform. Thus, the moment of inertia reduces to just the platform's moment of inertia.
Iplatform_new = Iplatform
The angular velocity of the platform when the child is at the center is given by:
wcenter = Lfinal / Iplatform_new
Since there is no external torque acting on the system, the final angular momentum is equal to the initial angular momentum. Therefore:
Lfinal = Iplatform * winitial
Substituting the given values:
Iplatform * winitial = Iplatform * wcenter
Simplifying:
wcenter = winitial
Therefore, the angular velocity of the platform when the child is at the center is equal to the initial angular velocity, which is 3.8 rad/s.
c) To find the change in kinetic energy when the child walks from the rim to the center of the platform, we can use the conservation of kinetic energy.
The initial kinetic energy is given by:
Kinitial = 0.5 * Iplatform * winitial^2
The final kinetic energy is given by:
Kfinal = 0.5 * Iplatform * wfinal^2
The change in kinetic energy is then:
ΔK = Kfinal - Kinitial
Substituting the given values:
ΔK = 0.5 * Iplatform * (wfinal^2 - winitial^2)
ΔK = 0.5 * 115 kg * (1.6 m)^2 * ((2.999 rad/s)^2 - (3.8 rad/s)^2)
Simplifying:
ΔK = 0.5 * 115 kg * (1.6 m)^2 * (8.994001 - 14.44) rad^2/s^2
Therefore, the change in kinetic energy when the child walks from the rim to the center of the platform is approximately -621.3 J (negative value indicating a decrease in kinetic energy).
To solve this problem, we need to 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, we can use the conservation of angular momentum. The initial angular momentum of the system is given by:
L_initial = I_platform * ω_initial
where I_platform is the moment of inertia of the platform and ω_initial is the initial angular speed.
The final angular momentum of the system can be expressed as:
L_final = (I_platform + I_child) * ω_final
where I_child is the moment of inertia of the child and ω_final is the final angular speed.
Since the platform is treated as a disk, its moment of inertia can be calculated using the formula:
I_platform = 0.5 * m * r^2
where m is the mass of the platform and r is its radius.
Substituting the given values, we have:
I_platform = 0.5 * 115 kg * (1.6 m)^2 = 147.2 kg·m^2
The moment of inertia of the child can be approximated as that of a point mass rotating about the rim of the disk, given by:
I_child = m * r^2
Substituting the given values, we have:
I_child = 40 kg * (1.6 m)^2 = 102.4 kg·m^2
Now, using the conservation of angular momentum:
L_initial = L_final
I_platform * ω_initial = (I_platform + I_child) * ω_final
Substituting the known values:
(147.2 kg·m^2) * (3.8 rad/s) = (147.2 kg·m^2 + 102.4 kg·m^2) * ω_final
Simplifying the equation, we find:
ω_final = (147.2 kg·m^2 * 3.8 rad/s) / (147.2 kg·m^2 + 102.4 kg·m^2)
With these values, we can calculate ω_final.
b) When the child moves to the center and stays there, the moment of inertia of the system changes. The final angular velocity in this case can be calculated using the conservation of angular momentum as:
L_initial = L_final
(I_platform + I_child) * ω_final_1 = I_platform * ω_final_2
Using the same values for I_platform and I_child as before:
(147.2 kg·m^2 + 102.4 kg·m^2) * ω_final_1 = 147.2 kg·m^2 * ω_final_2
Simplifying the equation, we find:
ω_final_2 = (249.6 kg·m^2 * ω_final_1) / (147.2 kg·m^2 + 102.4 kg·m^2)
With these values, we can calculate ω_final_2.
c) The change in kinetic energy when the child walks from the rim to the center of the platform can be calculated using the conservation of kinetic energy.
The initial kinetic energy of the system is given by:
KE_initial = 0.5 * I_platform * (ω_initial)^2
The final kinetic energy of the system when the child is at the center is:
KE_final = 0.5 * (I_platform + I_child) * (ω_final_2)^2
Substituting the known values, we have:
KE_initial = 0.5 * 147.2 kg·m^2 * (3.8 rad/s)^2
KE_final = 0.5 * (147.2 kg·m^2 + 102.4 kg·m^2) * (ω_final_2)^2
The change in kinetic energy is then given by:
ΔKE = KE_final - KE_initial
With these values, we can calculate ΔKE.