Question

Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ω0, but in opposite directions. The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity.

a) what is there final angular speed?
b) what is the change in the rotational kinetic energy?

Answer 1

Answer: A. {3*w¹}/5

B. ∆K.E = {-8*I*(w¹)²}/{25}

Explanation: Moment of inertia (I)for disk is {m*r²}/2.

Where r= radius of disk.

For large disk with r= 2r

I1= {m*(2r)²}=2*m*r².

For small disk

radius = r

Is= {m* r²}/2

Angular momentum(Am¹) for large

disk = I1*w¹ that is moment of inertia * angular speed.

Am¹= 2*m*r²*w¹.

Also, Angular momentum of small disk(Ams) is

Ams= - {m*r²*w¹}/2. The negative sign due to the fact from the question that it is rotating in opposite direction.

Am¹ + Ams = 2*m*r²w¹ - {m*r²*w¹}/2

= {3*m*r²*w¹}/2

From conservation of momentum.

Initial momentum = final momentum

{3*m*r²*w¹}/2 = {I1 +Is}*Wf

Wf = final angular speed.

I1 + Is = 2*m*r² + {m*r²}/2 = {5*m*r²}/2

Substituting into the conservation equation and cancelling out m*r²

We have that

Wf = {3*w¹}/5 . Which is the final angular speed.

B.

Kinectic energy (K.E) of a rotating body like the disk is = {moment of inertia(I) * {angular speed}²}/2

K.E final= {I * (Wf)²}/2= {I*{(3*w¹)/5}²}/2

K.E final = {I* 9*(w¹)²}/50.

Also, K.E initial = {I* (w¹)²}/2

Where w¹ is the initial angular speed.

But change in rotational K.E (∆K.E) = K E final - K.E initial.

∆K.E = {{9*I*(w¹)²}/50} - {{I*(w¹)²}/2}

Find the L..c.m of 2 and 50 first and simplifyng the fraction we have that,

Change in rotational K.E

= -{8*I*(w¹)²}/25 the negative sign indicates that there is a loss in kinetic energy.