A flywheel with a moment of inertia of 1.00kg m^2 is rotating at 100 rpm. What torque is required to bring it to rest in 10.0 revolutions?

average rpm during halt = 50 rev/min

time to halt = 10 rev/50 rev/min
= .2 min or 12 seconds

omega at start = 100 rev/min*2pi rad/rev * 1 min/60s

= 100 *2pi/60 radians/second

so alpha = d omega/dt
= - 10pi/3 / 12 radians/s^2
and
Torque = I * alpha

To find the torque required to bring the flywheel to rest, we need to use the principle of conservation of angular momentum.

The formula for angular momentum is:

L = Iω,

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

In this case, the initial angular momentum (L1) of the flywheel is given by:

L1 = I1ω1,

where I1 = 1.00 kg m^2 is the moment of inertia and ω1 = 2π(100/60) rad/s is the initial angular velocity (converted from rpm to rad/s).

To bring the flywheel to rest, its final angular velocity (ω2) will be zero. The final angular momentum (L2) will then be:

L2 = I2ω2 = 0,

where I2 = 1.00 kg m^2 is the same moment of inertia.

The change in angular momentum (ΔL) is given by:

ΔL = L2 - L1 = 0 - I1ω1.

Since the torque (τ) is equal to the rate of change of angular momentum, we can use the equation:

τ = ΔL / Δt,

where Δt is the time interval taken to bring the flywheel to rest. In this case, we're given that the flywheel comes to rest in 10.0 revolutions.

Since one revolution is equal to 2π radians, the time interval can be calculated as:

Δt = (10.0 revolutions / (2π(100/60) rad/s) = (10.0 * 2π / (2π(100/60))) s.

Now we can substitute the values into the equation:

τ = ΔL / Δt = (0 - I1ω1) / Δt.

Plug in the values:

τ = (0 - 1.00 kg m^2 * 2π(100/60) rad/s) / ((10.0 * 2π) / (2π(100/60))) s.

Calculating this expression will give you the torque required to bring the flywheel to rest.