The figure below shows three rotating, uniform disks that are coupled by belts.

Question

The figure below 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 0.5000R; disk B has radius 0.2250R; and disk C has radius 2.250R. 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?

Explanation / Answer

1: we know the tangential velocity of the tape because of its length and its duration. Assuming that equal distances on the tape correspond to equal times, it is moving at a constant tangential velocity.

v_tan = d/t
note, we convert t to 8280 seconds

Assume each reel has the same inner radius (Ri). Find the disk area occupied by the full reel and then distribute it such that both reels have the same outer radius:

Ro1 is the initial outer radius. Ro2 is the reel balanced outer radius

A_tape_full = Pi*(Ro1^2 - Ri^2)
2*A_tape_balance = A_tape_full, because the tape is balanced among the two reels

A_tape_balance = Pi*(Ro2^2 - Ri^2)

2*Pi*(Ro2^2 - Ri^2) = Pi*(Ro1^2 - Ri^2)

Solve for Ro2:
Ro2 = 1/2*sqrt(2*Ri^2 + 2*Ro1^2)

Now that we know Ro2, find corresponding angular velocity:
omega2 = v_tan/Ro2

Substitute:
omega2 = 2*d/(t*sqrt(2*Ri^2 + 2*Ro1^2))

Plug in data:
d:=196; t:=8280; Ri:=0.013; Ro1:=0.047;

Result:
omega2 = 0.6865 rad/sec

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2: Use kinematics to find the angular velocity as a function of time. Correspond the angular velocity and angular acceleration to tangential linear versions. Find the radial acceleration by using the centripetal acceleration formula.

omega = alpha*t

v = omega*R
a_tan = alpha*R

a_rad = v^2/R

Condition:
v^2/R = alpha*R

Substitute:
(omega*R)^2/R = alpha*R
(alpha*t*R)^2/R = alpha*R

Solve for t:
t = 1/sqrt(alpha)
notice this doesn't depend on R?

Plug in alpha = 1 rad/s^2

Result:
t = 1 second

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3:
A) radius of gyration of a solid disk about central axis:
I = 1/2*m*R^2
K^2 = I/m
K^2 = 1/2*R^2

K = R/sqrt(2)

R:=6.05 meters

Result:
K = 4.278 meters

B) uniform rod about a central axis

I = 1/12*m*L^2

K^2 = I/m

K^2 = 1/12*L^2

K = L/sqrt(12)
K = L/(2*sqrt(3))

L:= 7.69 meters

Result:
K = 2.220 meters

C) solid (assumed uniform) sphere about a central axis

I = 2/5 * m*R^2

K^2 = I/m
K^2 = 2/5*R^2

K = R*sqrt(2)/sqrt(5)

R:= 6.05 meters;

Result:
K = 3.826 meters;

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4: lug nut force. I assume 57 degrees is the angle from Parallel.

tau = L*F*sin(theta)

Solve for F:
F = tau/(L*sin(theta))

data:
tau := 72 N*m; L:=0.22 m; theta:=57 deg;

Result:
F = 390.2 Newtons

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5: The ball both slides and spins so that both kinetic energies are equal

KE_rot = KE_trans

KE_rot = 1/2*I*omega^2
KE_trans = 1/2*m*v^2

I = 0.36*m*R^2

1/2*m*v^2 = 1/2*(0.36*m*R^2)*omega^2

simplify:
v^2 = 0.36*(R*omega)^2

Find ratio of center of mass speed (v) to tangential speed due to rotation (omega*R)

Solve for v/(omega*R)

(v/(omega*R))^2 = 0.36

Result:
v/(omega*R) = 0.6

Ratio: 3 to 5

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