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Gnom [1K]
3 years ago
11

A carnival merry-go-round has a large disk-shaped platform of mass 120 kg that can rotate about a center axle. A 60-kg student s

tands at rest at the edge of the platform 4.0 m from its center. The platform is also at rest. The student starts running clockwise around the edge of the platform and attains a speed of 2.0 m/s relative to the ground.
(a) determine the rotational velocity of the platform.
(b) determine the change of kinetic enrgy of the system consisting of the platform and the student.
Physics
1 answer:
olasank [31]3 years ago
7 0

Answer

given,  

radius of merry - go - round = 4 m  

mass of the disk = 120 kg  

speed of the merry- go-round = 0 rad/s

speed = 2 m/s  

mass of student = 60 kg  

I_{disk} = \dfrac{1}{2}MR^2  

I_{disk} = \dfrac{1}{2}\times 120 \times 4^2  

I_{disk} = 960 kg.m^2  

initial angular momentum of the system  

L_i = I\omega_i + mvR  

L_i =960 \times 0 + 60 \times 2 \times 4  

L_i =480\ kg.m^2/s  

final angular momentum of the system    

L_f = (960 + 60\times 4^2)\omega_{f}    

from conservation of angular momentum  

L_i = L_f  

480 = (1920)\omega_{f} 

 now,

change in kinetic energy of the system

initially the merry-go-round was on rest KE_i = 0

\Delta KE = KE_f- KE_i

\Delta KE = (\dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2)- 0

\Delta KE = \dfrac{1}{2}\times 60 \times 2^2 + \dfrac{1}{2}\times 960 \times( {\dfrac{v}{r})^2

\Delta KE =120 + 480 \times \dfrac{4}{16}

\Delta KE = 240\ J

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