A 35-kg boy running at 3.9 m/s jumps tangentially onto a small circular merry-go

Question

A 35-kg boy running at 3.9 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 0.80 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.

Part B

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

Part C

Find the change in the boy's kinetic energy.

Part D

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

Explanation / Answer

apply angular momentum conservation

Li = Lf

35 * 3.9* 2 + 2*10^2 * 0.80 = ( 35 * 2^2 +2*10^2) *w

solve for w

w= 1.2535 rad/s

change in KE of system = KEf - KEi

= 0.5 (  ( 35 * 2^2 +2*10^2) *1.2735^2 - 0.5 *35*3.9^2 - 0.5* 2*10^2*0.80^2

= 54.46 J

change in the boy's kinetic energy

v = r*w = 2* 1.25

= 0.5*35*3.9^2 - 0.5 * 35 * ( 2* 1.2535 )^2

= 156.1866 J

change in the kinetic energy of the merry-go-round. = 0.5 * 2*10^2 *0.80^2 - 0.5*2*10^2 * 1.2535^2 = 93.126 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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