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A man of mass m stands on a flat horizontal disc (mass M, radius R) near its edge. The disc is free to rotate without friction about its central axis. At a certain instant the man begins to walk...

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A man of mass m stands on a flat horizontal disc (mass M, radius R) near its edge. The disc is free to rotate without friction about its central axis. At a certain instant the man begins to walk around the disc with a constant velocity v with respect to the ground.

If his feet do not slip on the disc, how long will it take him to return to the same point on the disc? Express your answer as a function of R, M, m and v

Once he gets to the starting point, he stops walking. How long after he stops walking will it take the disc to stop turning?Explain

The disc and the man are both stationary. If he walks in towards the centre of the disc, with a velocity u, how fast will the disc turn?Explain.

The man again stands at the edge of the disc at rest. A ball of mass m/4 is thrown to him with velocity v tangent to the rim of the disc. He catches the ball. How fast does the disc turn? Express your answer as a function of R, M, m and v.

Answered Same Day Dec 26, 2021

Solution

Robert answered on Dec 26 2021
119 Votes
SOLUTION

Man of mass m
M
Disc of radius R
R
(a)
Let’s assume beforehand that the man is starting to walk around the disc in the anti-clockwise
direction.
From an inertial frame of system (ground), we see man+wheel as a whole, so after the man has
started walking let’s assume that the wheel also started rotating in the anti-clockwise direction
with an angular velocity ω.
Velocity of man w.r.t (with respect to) ground = v
Conserving angular momentum before and after the man starts to walk on the disc
Initial angular momentum = 0
Final angular momentum with respect to ground = m*v*R + I*ω
Here, I = angular momentum of disc = (M*R2 )/2
So, 0 = m*v*R + ω*(M*R2 )/2

ω = - (2*m*v/ (M*R)) , Note that the negative sign...
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