merry-go-round is a playground ride that consists of a large disk mounted to tha

Question

merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R 1.3 meters, and a mass M 261 kg. A small boy of mass," 49 kg runs tangentially to the merry-go-round at a speed of v 2.1 m/s, and jumps on Randomized Variables R=13meters M-261 kg m=49kg v2.1 m/s Ctheexpertta.com Part (a) Calculate the moment of inertia of the merry-go-round, in kg m2 Numeric : A numeric value is expected and not an expression. Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round Numeric : A numeric value is expected and not an expression. Part (c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy Numeric :A numerie value is expected and not an expression. Part (d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry go-round when the boy is half way between the edge and the center of the merry go round? Numerie : A numeric value is expected and not an expression. Part (e) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the the boy is at the center of the merry go round? Numeric :A numeric value is expected and not an expression. 24 merry-go-round when Part (0 Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed jump off in radians/second of the merry-go-round after the boy jumps off? Numeric : A numeric value is expected and not an expression. Dana 3 of

Explanation / Answer

moment of inertia I = (1/2)*M*R^2

I = (1/2)*261*1.3^2 = 220.545 kg m^2

===================================

part(b)

w1 = v/R = 2.1/1.3 = 1.62 rad/s

part(c)

angular momentum before jumping

Li = m*v*R

angular momentum after jumping on to the merrygo round

Lf = ( I + mR^2)*w2

from conervation of angular momentum

Lf = Li

( 220.54 + 49*1.3^2)*w2 = 49*2.1*1.3

angular speed w2 = 0.44 rad/s

=================================

part(d)

after crawling to half way

final angular momentum Lf = (I + m(R/2)^2*w3

Li = Lf

m*v*r = (I + m(R/2)^2*w3

49*2.1*1.3 = ( 220.545 + 49*(1.3/2)^2 ) W3

w3 = 0.55 rad/s

===================

part(e)

after crawling to center

final angular momentum Lf = I *w4

Li = Lf

m*v*r = I*w4

49*2.1*1.3 = 220.545 *W4

w4 = 0.606 rad/s

=====================

part(f)

as the child crawls to edge

angular speed w2 = 0.44 rad/s

before jumping

angular momentum of child Lchild = m*R^2*w2

angular momentum of merrygo Lmerry = I*w2

Lbefore = m*R^2*w2 + I*w2

after jumping

angular momentum of child Lchild = 0

angular momentum of merrygo Lmerry = I*w5

Lafter = I*w5

from momentum conservation

Lafter = Lbefore

220.54*w5 = (49*1.3^2*0.44) + (220.54*0.44)

w5 = 0.605 rad/s

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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