A child of mass 30 kg stands beside a circular platform of mass 95 kg and radius 2.4 m spinning at 2.3 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?
Thank you so much!
You are welcome.
Remember that figure skater spinning slowly with arms extended, then suddenly speeding up when she pulls her arms in :)
To answer these questions, we need to understand the concept of conservation of angular momentum. Angular momentum is defined as the product of the moment of inertia and the angular velocity of an object. According to the conservation law, the total angular momentum of a system remains constant unless acted upon by an external torque.
a) To determine the new angular speed when the child steps on the rim of the platform, we can use the conservation of angular momentum. The initial angular momentum of the system is given by:
L_initial = I_initial * ω_initial
where L_initial is the initial angular momentum, I_initial is the initial moment of inertia, and ω_initial is the initial angular velocity.
Since the platform is initially spinning with an angular velocity of 2.3 rad/s, we can write:
L_initial = (I_platform + I_child) * ω_initial
where I_platform is the moment of inertia of the platform, I_child is the moment of inertia of the child, and ω_initial is the initial angular velocity.
The moment of inertia of a circular disk is given by:
I = (1/2) * m * r^2
where m is the mass of the object and r is the radius. Therefore, the moment of inertia of the platform is:
I_platform = (1/2) * 95 kg * (2.4 m)^2
Similarly, the moment of inertia of the child is:
I_child = (1/2) * 30 kg * (2.4 m)^2
Substituting the values, we can calculate the initial angular momentum:
L_initial = [(1/2) * 95 kg * (2.4 m)^2 + (1/2) * 30 kg * (2.4 m)^2] * 2.3 rad/s
Now, when the child steps on the rim, her moment of inertia changes because she becomes a part of the rotating system:
I_child_new = (1/2) * 30 kg * (0 m)^2
Since the child is at the rim, the new radius (distance from the center) is zero. As a result, the child's moment of inertia becomes zero.
Using the conservation of angular momentum, we can write:
L_initial = (I_platform + I_child_new) * ω_new
Solving for ω_new (the final angular velocity), we get:
ω_new = L_initial / (I_platform + I_child_new)
By plugging in the values, we can find the new angular speed.
b) When the child walks towards the center and reaches the center of the platform, the radius from the center becomes zero. Therefore, the moment of inertia of the child becomes zero. At this point, the child's contribution to the angular momentum becomes zero, and only the platform's moment of inertia affects the system's angular velocity.
Hence, the angular velocity of the platform when the child is at the center is the same as the new angular velocity calculated in part a).
c) The change in kinetic energy when the child walks from the rim to the center of the platform can be determined using the conservation of angular momentum. The initial kinetic energy of the system is given by:
KE_initial = (1/2) * (I_platform * ω_initial^2 + I_child * ω_initial^2)
Similarly, the final kinetic energy of the system is given by:
KE_final = (1/2) * (I_platform * ω_new^2 + I_child_new * ω_new^2)
The change in kinetic energy is obtained by subtracting the initial kinetic energy from the final kinetic energy:
ΔKE = KE_final - KE_initial
By substituting the values, we can calculate the change in kinetic energy when the child walks from the rim to the center of the platform.
Ip = I of platform =(1/2)(95)(2.4^2)
Angular momentum = Ip(2.3)
IT STAYS THAT FOREVER
New In = Ip +30*(2.4^2)
In omega = Ip (2.3)
omega = (Ip/In)(2.3) (Part a)
b)
Her I = mr^2 is zero
BACK to original I and 2.3 rad/s
c)
(1/2) In omega^2 - (1/2)Ip(2.3^2)