A 40 kg boy running at 4.0 m/s jumps tangentially onto a small stationary circul
ID: 1507691 • Letter: A
Question
A 40 kg boy running at 4.0 m/s jumps tangentially onto a small stationary circular merry - go - round of radius 2.0 m and rotational inertia 20 kg. m2 pivoting on a frictionless bearing on its centrl shaft.
(a) Determine the rotational velocity of the merry - go - round after the boy jumps on it.
(b) Find the chance in kinetic energy of the system consisting of the boy and the merry - go - round.
(c) Find the change in the boy's kinetic energy.
(d) Find the change in the kinetic energy of the merry - go - ground.
(e) Compare the kinetic energy changes in parts (b) through (d).
Explanation / Answer
a) Let’s determine the boy’s initial momentum.
M = 40 * 4.0 = 160.0
Let’s determine the boy’s initial kinetic energy.
KE = ½ * 40 * 4.0^2 = 320 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 = 40 * 2^2 = 160
To determine his angular velocity, divide his velocity by 2.
= 2.0 rad/s
Initial angular momentum = 160 * 2.0 = 320
Total moment of inertia = 160 + 20 = 180
Angular momentum = 180 * = 320
180 * = 320
= 1.78 rad/s
Rotational KE = ½ * I * ^2 = ½ * 180 * (320/180)^2
Rotational KE = 284.44 J.
To determine the total decrease of kinetic energy, subtract his number from the boy’s initial kinetic energy.
Decrease of KE = 320 – (½ * 320^2/180) = 35.56 J
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