Can anyone help me out with this question, with steps?

Question

A uniform 2 kg disk of radius 25 cm is rotating around its central axis at an angular

speed of 20 rad/s. At time t=0, a man begins to slow it at a uniform rate until it stops at t = 8 s.

(a) By time t = 5 s, how much work had the man done?

(b) For full 8 s, at what average rate did the man do work?

Answer 1

(a) This problem can be solved by the relationship between Work and Kinetic Energy. For any object, the net work on the object equals the change in kinetic energy.

(1) W = KE_f - KE_i

For a rotating body, the kinetic energy depends on the angular velocity and the Moment of Inertia about its axis of rotation.

(2) KE = (1/2)*I*ω^2

For a disk, its Moment of Inertia depends on its mass and radius

(3) I_d = (1/2)*M*R^2

substituting equation (3) into (2), and then (2) into (1) gives

(4) KE = (1/2)*(1/2)*M*R^2*ω^2

(5) W = (1/2)*(1/2)*M*R^2*ω_f^2 - (1/2)*(1/2)*M*R^2*ω_i^2 = (1/4)*M*R^2*(ω_f^2-ω_i^2)

The mass and radius of the disk are given constants, and the initial angular velocity is also given. The final angular velocity needs to be found by determining by how much the disk has slowed after 5 seconds. For constant acceleration:

(6) ω_f = ω_i + αt (where t= 5 seconds)

The angular acceleration can be determined from the fact that the disk comes to a stop after 8 seconds

(7) α = (Δω)/(Δt) = (-ω_i)/(Δt) (where Δt = 8 seconds)

This gives α=-20/8 = -2.5 rad/s^2

Substituting into equation (6) gives

ω_f = ω_i + αt = 20+(-2.5)(5) = 7.5 rad/s

This value for the final angular speed can be substituted back into equation (5) to give

W = (1/4)*M*R^2*(ω_f^2-ω_i^2) = (1/4)*(2)*(0.25)^2*(7.5^2-20^2) = -10.74 J

The man does -10.74 J of work on the system, meaning that he removes 10.74 J of work.

(b) For this part, the same analysis can be done. However, since part b asks about the rate of work over the entire 8 seconds, the final velocity is not 7.5 rad/sec, but 0, because the disk comes to a stop at that time. Returning to equation (5)

W = (1/4)*M*R^2*(ω_f^2-ω_i^2) = (1/4)*(2)*(0.25)^2*(0^2-20^2) = -12.5 J

The average rate of work (Power) done over a time interval is the change in work over the change in time

P = (ΔW)/(Δt) = (-12.5)/(8) = -1.56 J/s = -1.56 W

The average rate of work done is -1.56 W.

Answer 2

Momentum inershiya of disk: I = 0.5mr² = 0.5·2kg·0.25²m² = 0.0625kgm².

(a) ω = ω₀ + εt; ε = (ω - ω₀)/t = (0 - 20rad/s)/s = - 2.5 rad/s².

at t = 5s ω = 20rad/s - 2.5rad/s²·5s = 7.5 rad/s

Work W is equel change of knetic energy of disk: W = 05I(ω² - ω₀²) = 0.5·0.0625kgm²(56.25 - 400)rad²/s² =

-10.74 J

(b) This is average power P = [0.50.0625kgm²(0 - 20²)kgm²rad²/s²]/8s= - 1.5625 W

Answer 3

The key to doing rotational problems is to understand the equations of translational mechanics and use the analogy to rotational mechanics:

a) The question is work done by the man. There are two approaches: Linear: W = Fd where d is the amount of edge of the disk that goes around in the first 5 seconds. The other is to use the analogous rotational equation W = τ Δθ with torque for force and change in angle (rads) for displacement. I think rotational is easier as you have to use rotational kinematic equation:

From mechanics (F=ma), you know that τ=Iα (have I = 1/2 Mr² for a disk) have to get the rotational acceleration from the kinematics:

v = v₀ + at or ω = ω₀ + αt use the 8 second data with ω(8s) = 0 to find α.

Now you can solve for the torque from the previous equation.

You need Δθ which can be found as you know 3/5 equation of motion variables: α,ω₀, and t = 5

d = v₀ t + 1/2 at² --> Δθ = ω₀t + 1/2αt²

Finally, you can find W = τΔθ (note α, τ, and W will be negative as they act against the positive rotation)

b) To solve this problems it is easier to solve this using energy: E_K,rot,intial = W (all of the kinetic energy of the disk was lost to frictional work (thermal energy) ) Total work = 1/2 I ω₀² and the average rate of work is just the (total work)/(total time)

Please consider a tutor.