A spinning flywheel has a rotational inertia I=400.0 kg metre squared.Its angular velocitydecreases from 20.0rad/secondsto zero in 300.0 seconds due to friction.What is the frictional torque acting?
goodness. See other post
torque = I(moment of inertia) times angular acceleration(change in angular velocity over time) change in angular velocity over time simply put is velocity final - velocity intial divided by total time that's all im doing for you,you should get it form that
To find the frictional torque acting on the spinning flywheel, we can use the equation:
ΔE = τθ
Where ΔE is the change in the rotational kinetic energy of the flywheel, τ is the torque acting on the flywheel, and θ is the angle through which the torque acts.
In this case, the flywheel starts with an initial angular velocity ω_i = 20.0 rad/s and ends with a final angular velocity of ω_f = 0. We also know that it takes 300.0 seconds for the flywheel to come to rest.
The change in angular velocity is given by:
Δω = ω_f - ω_i
Δω = 0 - 20.0 rad/s
Δω = -20.0 rad/s
Using the formula for rotational kinetic energy:
E = 1/2 I ω^2
The initial rotational kinetic energy is:
E_i = 1/2 I ω_i^2
E_i = 1/2 (400.0 kg*m^2) (20.0 rad/s)^2
E_i = 1/2 (400.0 kg*m^2) (400.0 rad^2/s^2)
E_i = 40,000.0 J
The final rotational kinetic energy is:
E_f = 1/2 I ω_f^2
E_f = 1/2 (400.0 kg*m^2) (0)^2
E_f = 1/2 * 0
E_f = 0 J
Therefore, the change in rotational kinetic energy is:
ΔE = E_f - E_i
ΔE = 0 J - 40,000.0 J
ΔE = -40,000.0 J
Since the flywheel is slowing down, the change in kinetic energy is negative.
Now, we can substitute the values into the formula ΔE = τθ:
-40,000.0 J = τ (300.0 s)
Solving for τ:
τ = ΔE / θ
τ = (-40,000.0 J) / (300.0 s)
τ = -133.33 N*m
The frictional torque acting on the spinning flywheel is -133.33 N*m. The negative sign indicates that the torque is opposing the motion and causing the flywheel to slow down.