The figure shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks A and C. Another belt runs around a central hub on disk A and the rim of disk B. The belts move smoothly without slippage on the rims and hub. Disk A has radius R; its hub has radius 0.494R; disk B has radius 0.216R; and disk C has radius 1.64R. Disks B and C have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk C to that of disk B?

To find the ratio of the magnitude of the angular momentum of disk C to that of disk B, we need to know their respective angular momenta. The angular momentum of an object can be calculated using the formula:

L = I * ω

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

The moment of inertia, I, for a uniform disk rotating about its central axis is given by:

I = (1/2) * m * r^2

Where m is the mass of the disk and r is its radius.

In this case, both disks B and C have the same density and thickness, so we can assume that their masses are proportional to their areas.

The area of a disk is given by:

A = π * r^2

We can assume a constant thickness for all the disks, which means the masses are proportional to the radii squared:

m_B = k * (0.216R)^2
m_C = k * (1.64R)^2

Where k is a constant of proportionality.

Now, let's calculate the angular velocity of each disk. From the problem statement, we know that the belts move smoothly without slippage on the rims and hub. This means that each belt is moving at the same linear speed.

The linear speed, v, of a point on the rim of a rotating disk is given by:

v = ω * r

Where ω is the angular velocity and r is the radius of the disk.

For belt 1 (around A and C), the linear speed is the same for both disks A and C:

v_1_A = v_1_C

ω_1_A * (R + 0.494R) = ω_1_C * (1.64R)

For belt 2 (around the hub of A and rim of B), the linear speed is also the same:

v_2_A = v_2_B

ω_2_A * 0.494R = ω_2_B * (0.216R + 0.494R)

Now, let's solve the equations to find the angular velocities of each disk.

Divide the equation for belt 1 by the equation for belt 2:

(ω_1_A * (R + 0.494R)) / (ω_2_A * 0.494R) = (ω_1_C * (1.64R)) / (ω_2_B * (0.216R + 0.494R))

Simplify:

((R + 0.494R) / (0.494R)) * ((0.216R + 0.494R) / (1.64R)) = (ω_1_C / ω_2_B)

Simplify further:

1.494 * 0.956 = (ω_1_C / ω_2_B)

Now, we have the ratio of the angular velocities of disk C to disk B.

The angular momentum of a disk can be calculated by multiplying the moment of inertia by the angular velocity.

L_B = I_B * ω_2_B
L_C = I_C * ω_1_C

Using the formulas for moment of inertia and the ratio of angular velocities, we can find:

L_C / L_B = (I_C * ω_1_C) / (I_B * ω_2_B)

Substituting the expressions for moment of inertia from earlier:

L_C / L_B = ((1/2) * m_C * (1.64R)^2 * ω_1_C) / ((1/2) * m_B * (0.216R)^2 * ω_2_B)

Cancel out the factors of 1/2:

L_C / L_B = (m_C * (1.64R)^2 * ω_1_C) / (m_B * (0.216R)^2 * ω_2_B)

Substitute the expressions for masses and the ratio of angular velocities:

L_C / L_B = ((k * (1.64R)^2) * ω_1_C) / ((k * (0.216R)^2) * ω_2_B)

Simplify further:

L_C / L_B = ((1.64R)^2 * ω_1_C) / ((0.216R)^2 * ω_2_B)

Note that the k constant cancels out.

L_C / L_B = (1.64^2 * R^2 * ω_1_C) / (0.216^2 * R^2 * ω_2_B)

Simplify:

L_C / L_B = (1.64^2 * ω_1_C) / (0.216^2 * ω_2_B)

Finally, we have the ratio of the magnitude of the angular momentum of disk C to that of disk B, which is:

L_C / L_B = (1.64^2 * ω_1_C) / (0.216^2 * ω_2_B)

Solving this equation will give you the desired ratio.

Honestly, I don't get the picture.