A 40-kg boy running at 4.0 m/s jumps tangentially onto a small circular merry-go

Question

A 40-kg boy running at 4.0 m/s jumps tangentially onto a small circular merry-go-round of radius 2.0 m and rotational inertia 20 kgm2pivoting on a frictionless bearing on its central shaft. Merry-go-round initially rotating at0.60 rad/s opposite the direction that the boy was running before he jumped on it.

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

B. Find the change 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-round.

Explanation / Answer

I = moment of inertia of merry-go-round = 20 kgm2

r = radius = 2 m

m = mass of boy = 40 kg

v = 4 m/s

wi = initial angular speed = 0.60 rad/s

A)

Using conservation of angular momentum

Iwi - mvr = (I + mr2) Wf

20 (0.6) - 40 x 4 x 2 = (20 + 40 (2)2) Wf

Wf = - 1.71 rad/s

B)

Change in KE = (0.5) I Wi2 + (0.5) m v2 - (0.5)(I + mr2) W2f

Change in KE = (0.5) 20 (0.6)2 + (0.5) (40) (4)2 - (0.5) (20 + 40 (2)2) (- 1.71)2 = 60.43 J

C)

change in Boy's KE = (0.5) m v2 - (0.5)(mr2) W2f = (0.5) (40) (4)2 - (0.5) (40 (2)2) (1.71)2 = 86.1 J

D)

Change in KE of mery-go round = (0.5) I Wi2 - (0.5)(I ) W2f = (0.5) 20 (0.6)2 - (0.5) (20 ) (- 1.71)2 = - 25.64

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