A 38-kg boy running at 3.7m/s jumps tangentially onto a small circular merry-go-

Question

A 38-kg boy running at 3.7m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 20 kg?m2pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at0.70rad/s opposite the direction that the boy was running before he jumped on it.

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

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

Find the change in the boy's kinetic energy.

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

Explanation / Answer

a) conservation of angular momentum

I w + m v r = ( I + mr^2) w

20*-0.7 + 38*2*3.7 = (20 + 38*2^2)*w

w=1.55

b)

dKE = KE final - KE initial = ( 0.5*(20+38*2^2)*1.55^2 ) - (0.5*20*0.7^2 + 0.5*38*3.7^2) = -58.4 J

c) dKE boy = 0.5*(38*2^2)*1.55^2 - 0.5*38*3.7^2= -77.52 J

d) dKE merry go round = 0.5*20*1.55^2 - 0.5*20*0.7^2= 19.13 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.