Kepler (mass M) is playing at the local park. Due to some fantastically convenie

Question

Kepler (mass M) is playing at the local park. Due to some fantastically convenient urban planning. there are two merry-go-rounds (MGR) right next to each other. Kepler starts by sitting on the edge of the small MGR (mass 2M). rotating with it at an angular velocity of wo.. The large MGR (mass 4M) is currently stationary. The MGRs can be modeled as solid disks. 2.) (3pts) Ii Kepler jumps off the small MGR at Velocity V. as shown in the drawing. what will the final angular velocity (we) of the small MGR be? Your answer should be in terms of V. R and wo only.

Explanation / Answer

By conservation of angular momentum, Ltot.i = Lkepler f + LMGR f

As: Ltot i = Itot * wo

and Itot = IMGR + IKepler

As the MGR is a solid disk, and Kepler is a particle,

Itot = 1/2 (2M) R2 + M R2

Itot = 2 M R2

Thus,

Ltot i = 2 M R2 * wo

Going back to conservation of angular momentum,

Ltot i = Lkepler f + LMGR f

2 M R2 * wo = Lkepler f + LMGR f

As

Lkepler f = M V R

and

LMGR f = IMGR w1 = M R2 w1

Then

2 M R2 wo = M V R + M R2 w1

Cancelling M R,

2 R wo = V + R w1

Solving for w1,

w1 = 2wo - V/R [ANSWER]

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