Question

Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia 3.76 kg·m2 about its central axis, is set spinning counterclockwise (which may be taken as the positive direction) at 436 rev/min. The second disk, with rotational inertia 9.20 kg·m2 about its central axis, is set spinning counterclockwise at 953 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at 953 rev/min, what are their (b) angular velocity (using the correct sign for direction) and (c) direction of rotation after they couple together?

Answer 1

A) ω = 13.38 rev/s

B) ω = 9.167 rev/s

C) In clockwise direction

Step-by-step explanation:

We are given;

Rotational Inertia of first disk; I_1 = 3.76 kg·m²

Angular velocity of first disk; ω_1 = 436 rev/min = 7.267 rev/s

Rotational Inertia of second disk; I_2 = 9.2 kg·m²

Angular velocity of second disk; ω_2 = 953 rev/min = 15.883 rev/s

Total rotational inertia is;

I_total = I_1 + I_2

I_total = 3.76 + 9.2 = 12.96 kg·m²

Now; total angular momentum will be;

L_total = L_1 + L_2

Where L_1 is angular momentum of first disk and L_2 is angular momentum of second disk.

Thus;

I_total•ω = I_1•ω_1 + I_2•ω_2

Plugging relevant values in, we can find their angular speed after coupling which is ω.

Thus;

12.96ω = (3.76 x 7.267) + (9.2 x 15.883)

12.96ω = 173.44752

ω = 173.44752/12.96

ω = 13.38 rev/s

B) since second disk is now spinning clockwise, thus;

I_total•ω = I_1•ω_1 - I_2•ω_2

12.96ω = (3.76 x 7.267) - (9.2 x 15.883)

12.96ω = -118.8

ω = -118.8/12.96

ω = -9.167 rev/s

The negative sign tells us that it is clockwise.

So we would say ω = 9.167 rev/s in clockwise direction