A hollow sphere of radius 25 cm is rotating at 13 rad/ sec about an axis that passes through its center. The mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?
Just to comment, I'm not trying to hw dump like the person underneath me. If someone can help me get started I can usually finish
The work required to get it to stop is minus the initial kinetic energy,
-(1/2) I w^2
w = 13 rad/s in this case.
I is the moment of inertia, which for a hollow sphere of mass M and radius R is
I = (2/3)M*R^2
To find the amount of work required to bring the sphere to a stop, we need to calculate the rotational kinetic energy of the sphere in its initial state and in its final state at rest. The work done will be equal to the change in kinetic energy.
The rotational kinetic energy of a rotating object can be calculated using the formula:
KE = (1/2) * I * ω^2
Where:
- KE is the rotational kinetic energy
- I is the moment of inertia of the object
- ω is the angular velocity of the object
The moment of inertia of a hollow sphere is given by the formula:
I = (2/3) * m * r^2
Where:
- m is the mass of the sphere
- r is the radius of the sphere
In the initial state, the sphere is rotating at an angular velocity of 13 rad/sec. Therefore, we can calculate the initial rotational kinetic energy (KE_initial) using the given values:
r = 25 cm = 0.25 m (radius in meters)
m = 3.8 kg
I = (2/3) * m * r^2
I_initial = (2/3) * 3.8 kg * (0.25 m)^2
ω_initial = 13 rad/sec
KE_initial = (1/2) * I_initial * ω_initial^2
Now, we need to find the final rotational kinetic energy (KE_final) when the sphere comes to a stop. At rest, the angular velocity (ω_final) will be 0.
KE_final = (1/2) * I_initial * ω_final^2
The work done (W) to bring the sphere to a stop is given by the change in kinetic energy:
W = KE_final - KE_initial
Substituting the values, we can calculate the work done.