A disk is rotating at 50rps (revolution per second) about its fixed axis (delta) with respect to which its moment of inertia is J= 2×10^-2 kg.m^2 A braking couple of moment -0.5N.m is applied to the disk (the sense of rotation of the disk is taken as positive ). The disk will stop after a duration t equals to what?

wf=wi+alpha*t

but alpha (acceleration)=force/I

so solve for time t.
2PI*0=2PI*50-.5/2E-2 * t

To find the duration (t) it takes for the disk to stop rotating, we can use the rotational analog of Newton's second law of motion, which states that the torque (τ) applied to an object is equal to the moment of inertia (J) times the angular acceleration (α).

Since the braking couple of moment is given as -0.5 N.m (negative sign indicates opposite direction of rotation), we can use this value for the torque (τ). The moment of inertia (J) is given as 2×10^-2 kg.m^2.

The angular acceleration (α) is related to the angular velocity (ω) and the duration (t) by the equation:

α = (change in ω) / t

As the disk stops rotating, its final angular velocity (ωf) will be zero, and the initial angular velocity (ωi) is given as 50 rps (revolutions per second).

Let's convert the initial angular velocity (ωi) to radians per second (rad/s) for uniformity:

ωi = 50 rps × 2π radians per revolution ≈ 314.16 rad/s

Now, we can rearrange the equation α = (change in ω) / t to solve for t:

α = (ωf - ωi) / t

Since ωf is 0, the equation simplifies to:

0 = (0 - 314.16) / t

Solving for t:

0 = -314.16 / t

Multiply both sides by t:

0 = -314.16

Since the left side is always zero, the equation is satisfied for any value of t. This means that the duration (t) it takes for the disk to stop rotating is undefined.