A 37-kg boy running at 4.5 m/s jumps tangentially onto a small circular merry-go

Question

A 37-kg boy running at 4.5 m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 2.0×102 kgm2 pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at 1.0 rad/s in the same direction that the boy is running.

Part A

Determine the rotational speed of the merry-go-round after the boy jumps on it.

Express your answer to two significant figures and include the appropriate units.

Part B

Find the change in kinetic energy of the system consisting of the boy and the merry-go-round.

Express your answer to two significant figure and include the appropriate units.

Part C

Find the change in the boy's kinetic energy.

Express your answer to two significant figure and include the appropriate units.

Part D

Find the change in the kinetic energy of the merry-go-round.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

by conservation of momentum

initial momentum = final momentum

37 * 4.5 + 2 * 10^2 * 1 = (2 * 10^2 + 37 * 2^2) * w

rotational speed of the merry-go-round after the boy jumps on it w = 1.0531 rad/sec

change in kinetic energy of system = initial kinetic energy - final kinetic energy

change in kinetic energy of system = 0.5 * 37 * 4.5^2 + 0.5 * 2 * 10^2 * 1^2 - 0.5 * (2 * 10^2 + 37 * 2^2) * 1.0531^2

change in kinetic energy of system = 281.65558786 J

change in boy's kinetic energy = initial kinetic energy - final kinetic energy

change in boy's kinetic energy = 0.5 * 37 * 4.5^2 - 0.5 * 37 * 2^2 * 1.0531^2

change in boy's kinetic energy = 292.55754886 J

change in kinetic energy of merry go round = initial kinetic energy - final kinetic energy

change in kinetic energy of merry go round = 0.5 * 2 * 10^2 * 1^2 - 0.5 * 2 * 10^2 * 1.0531^2

change in kinetic energy of merry go round = -10.901961 J

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