A hanging mass is connected to a spool by a string going over a massless, fricti

Question

A hanging mass is connected to a spool by a string going over a massless, frictionless pulley. There is a solid circular disk on top of the spool. The hanging mass is released from rest. Due to it falling, the spool with the disk on it rotates. When the mass reaches the floor, it disconnects from the spool, which continues to rotate. A uniform ring is then very gently placed on top of the solid disk (the impact of placing the ring on has no effect on the system, just it's mass, size, and shape matters).

Variables:
g = gravity
h = height above the floor the hanging mass is released from
mh = mass of the hanging mass
rs = radius of the spool
md = mass of the solid disk
rd = radius of the solid disk
mr = mass of the uniform ring
ror = outer radius of the uniform ring
rir = inner radius of the uniform ring

1. Determine net torque acting on the pulley

2. Use Newton's 2nd Law for rotational motion to determine a relation between the angular acceleration and the moment of inertia of the rotating system

3. Using conservation of energy, determine the linear speed that the hanging mass will have when it hits the floor

4. When the hanging mass hits the floor, the circular diskwill rotate at constant angular speed. Use conservation of mechanical energy to calculate the angular speed of the disk just before it hits the floor

5. A uniform ring is gently dropped on the circular disk (after it reaches constant angular speed). Use conservation of angular momentum to find an expression for the moment of inertia of the ring in terms of i (found in step 3), f , and Idisc

6. Write the theoretical expression for calculation of the moment of inertia of a circular disk and uniform ring

7. Obtain an expression for the final angular speed.

8. Was the collision of ring with the rotating disk elastic? If not what percentage of the rotational kinetic energy was lost during collision?

Explanation / Answer

PE lost by hanging mass = (Mh)gh

1. Tension in string = T
(Mh)g - T = ((Mh))a
T*rs = I(alpha) = (0.5*Md*(rd)^2)a/rs
so (Mh)g - 0.5Mdrd^2*a/rs^2 = Mh * a
a = (Mh)g/(0.5Md*(rd/rs)^2 + Mh)
Torque on pulley = T*rs = 0.5*Md*(rd)^2*(Mh)g/rs*(0.5Md*(rd/rs)^2 + Mh)
2. Torque, T(rs) = I*alpha = (Mh)g/(0.5Md*(rd/rs)^2 + Mh)
3. linear speed = v
ke = 0.5(Mh)v^2
pe lost = (Mh)gh
ke of rotating spool = 0.5(Iw^2) = 0.5(0.5*Md*rd^2)(v/rs)^2
by conservation of energy
(Mh)gh = 0.5(Mh)v^2 + 0.5(0.5*Md*rd^2)(v/rs)^2
v = sqroot((Mh)gh/(0.5Mh + 0.25Md*(rd/rs)^2))
4. w = v/rs = sqroot((Mh)gh/(0.5Mh + 0.25Md*(rd/rs)^2))/rs
5. From conservation of angular momentum
Iiwi = Ifwf
(0.5Md*rd^2)*sqroot((Mh)gh/(0.5Mh + 0.25Md*(rd/rs)^2))/rs = (0.5Md*rd^2 + 0.5*Mr*(ror^2 - rin^2))wf

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