A wheel with rotational inertia I initially rotates with a angular speed ωi . A torque is applied, speeding the wheel up to three times the angular speed in time Δt .
A) Find the magnitude of the applied torque.
B) Find the work done on the wheel during this time.
To solve this problem, we need to use the formulas and principles of rotational motion.
A) To find the magnitude of the applied torque, we can use Newton's second law for rotational motion:
τ = Iα
Where τ represents the torque applied, I represents the rotational inertia, and α represents the angular acceleration.
We can rearrange the equation to solve for torque:
τ = I(ωf - ωi)/Δt
Given that the final angular speed (ωf) is three times the initial angular speed (ωi), we have:
τ = I(3ωi - ωi)/Δt
= I(2ωi)/Δt
= 2Iωi/Δt
Therefore, the magnitude of the applied torque is 2Iωi/Δt.
B) To find the work done on the wheel during this time, we can use the work-energy theorem for rotational motion:
W = ΔKE
Where W represents the work done, and ΔKE represents the change in rotational kinetic energy.
The change in rotational kinetic energy can be expressed as:
ΔKE = 1/2 I ωf^2 - 1/2 I ωi^2
Given that the final angular speed (ωf) is three times the initial angular speed (ωi), we have:
ΔKE = 1/2 I (9ωi^2) - 1/2 I ωi^2
= 4Iωi^2
Therefore, the work done on the wheel during this time is 4Iωi^2.