A 45- kg boy running at 3.8 m/s jumps tangentially onto a small circular merry-g

Question

A 45- kg boy running at 3.8 m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 2.0×102 kg m 2 pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at 0.90 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

Let’s determine the boy’s initial momentum.

M = 45 * 3.8 = 171
Let’s determine the boy’s initial kinetic energy.

KE = ½ * 45 * 3.8^2 = 324.9 J

Since the boy is on the edge of the merry-go-round, let’s use the following equation to determine his moment of inertia.

I = m * r^2 = 45 * 2^2 = 180
To determine his angular velocity, divide his velocity by 2.
= 1.9 rad/s

Initial angular momentum = 180 * 1.9 = 342
Total moment of inertia = 180 + 200 = 380
Angular momentum = 380 * = 342

380 * = 342
= 342 ÷ 380
This is approximately 0.90 rad/s.

Rotational KE = ½ * I * ^2 = ½ * 380 * (342 ÷ 380)^2
Rotational KE = ½ * 342^2/380
This is approximately 153.9 J.

To determine the total decrease of kinetic energy, subtract his number from the boy’s initial kinetic energy.

Decrease of KE = 324.9 – (½ * 342^2/380)
= 171 J.

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