A 41-kg boy running at 4.5 m/s jumps tangentially onto a small stationary circul

Question

A 41-kg boy running at 4.5 m/s jumps tangentially onto a small stationary circular merry-go-round of radius 2.0 m and rotational inertia2.0×102 kgm2 pivoting on a frictionless bearing on its central shaft.

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.

SubmitMy AnswersGive Up

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.

SubmitMy AnswersGive Up

Part C

Find the change in the boy's kinetic energy.

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

SubmitMy AnswersGive Up

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.

SubmitMy AnswersGive Up

Explanation / Answer

A) let w is the angular speed afetr the boy jumps onto merry-go-round.

Apply conservation of angular momentum

initial angular momentum = final angular momentum

m*v*R = (I+m*R^2)*w

==> w = m*v*R/(I+m*R^2)

= 41*4.5*2/(2*10^2+41*2^2)

= 1.01 rad/s

B) Ki = 0.5*m*v^2

= 0.5*41*4.5^2

= 415.1 J

Kf = 0.5*(I+m*R^2)*w^2

= 0.5*(2*10^2 + 41*2^2)*1.01^2

= 185.7 J

delta_K = Kf-ki

= 185.7 - 415.1

= -229.4 J

c) delta_k_boy = 0.5*(m*R^2)*w^2 - 0.5*m*v^2

= 0.5*(41*2^2)*1.01^2 - 0.5*41*4.5^2

= -331.5 J

D) delta_k_merry-go-round = 0.5*I*w^2 - 0

= 0.5*2*10^2*1.01^2 - 0

= 102.1 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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