How Much Power Is Needed to Stop a Rotating Wheel?

[BJN153]

Hello, please help me with this problem. thanks

A 22.0 kg wheel, essentially a thin hoop with radius 1.00 m, is rotating at 230 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it? What is the required power?

 

[xanthym]

Hello, please help me with this problem. thanks

A 22.0 kg wheel, essentially a thin hoop with radius 1.00 m, is rotating at 230 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it? What is the required power?

From problem statement:
{Wheel Mass} = M = (22.0 kg)
{Wheel Radius} = R = (1.00 m)
(Wheel Rotation Frequency} = f =(230 rpm) = (3.83333 rev/sec)
{Wheel Angular Speed} = ω = 2*π*f = 2*π*(3.83333 rev/sec) = (24.0855 radians/sec)
{Thin Hoop Moment of Inertia} = I = M*R² = (22.0 kg)*(1.00 m)² = (22 kg*m²)
{Required Stopping Time} = ΔT = (10 sec)

The wheel's Rotational Kinetic Energy is given by:
{Wheel Rotational Kinetic Energy} = (1/2)*I*ω² =
= (1/2)*(22 kg*m²)*(24.0855 radians/sec)² =
= (6381.22 Joules)
::: ⇒ {Work Required To Stop} = (6381.22 Joules)

{Ave Power Required To Stop} = {Work Required To Stop}/ΔT =
= (6381.22 Joules)/(10 sec) =
= (638.122 Watts)


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[thepatientmental]

The required power can be calculated by dividing the work done by the time taken. In this case, the work done is equal to the change in kinetic energy, which is equal to 1/2 * mass * (final velocity)^2. Since the wheel must be brought to a stop, the final velocity is 0. Therefore, the work done is 1/2 * 22.0 kg * (0 m/s)^2 = 0 J.

To calculate the required power, we need to divide the work done by the time taken. The time taken is given as 10 s. Therefore, the required power is 0 J / 10 s = 0 watts. This means that no power is required to stop the wheel in 10 seconds.

However, if we want to stop the wheel in a shorter period of time, let's say 5 seconds, then the required power would be 0 J / 5 s = 0 watts. This shows that the required power is dependent on the time taken to stop the wheel. The shorter the time, the higher the required power.

I hope this helps with your problem. Let me know if you have any further questions. Good luck!