Question

A 90-kg merry-go-round of radius 2.0 m is spinning at a constant speed of 20 revolutions per minute. A kid standing on the ground decides to bring the merry-go-round to rest and applies a force of 10.0 N tangentially to the merry-go-round. If the merry-go-round is modelled as a solid cylinder, (A) calculate its moment of inertia. (B) What is the angular acceleration of the merry-go-round? (C) How many revolutions will the merry-go-round complete until it finally stops?

1) [25 pts] A 90-kg merry-go-round of radius 2.0 m is spinning at a constant speed of 20 revolutions per minute. A kid standing on the ground decides to bring the merry-go-round to rest and applies a force of 10.0 N tangentially to the merry-go-round. If the merry-go-round is modelled as a solid cylinder, (A) calculate its moment of inertia. (B) What is the angular acceleration of the merry-go-round? (C) How many revolutions will the merry-go-round complete until it finally stops?

Answer 1

Mass of merry go around = M=90 kg

radius of merry go round = R = 2 m

Tangential force applied by the kid to stop the merry go round = F = 10 N

Initial angular speed of merry go rounbd = w = 20 rpm = 20*2*pi/60 = 2.09 rad/s

1) Mass moment of inertia of solid cylinder is given by I = 0.5*M*R2

Moment of inertia of merry go round = I=0.5*90*22 = 180 kg-m2

2) Tangential force applied by kid = F = 10 N

radius of merry go round = R =2 m

Torque applied by kid to stop the merry go round= T = F*R =10*2 = 20 Nm

let angular acceleration be $\alpha$

T= I*$\alpha$

Or, 20 = 180*$\alpha$

$\alpha$ = 20/180 = 0.11 rad/s2

Since it is decelerating the merry go round, the angular acceleration = -0.11 rad/s2

3) Initial angular velocity = w= 2.09 rad/s

angular acceleration = $\alpha$ = -0.11 rad/s2

final angula velocity = wf = 0 rad/s

Lte total angular displacement of merry go round before coming to stop be $\theta$

wf2 = w2 + 2*$\alpha$*$\theta$

02 = 2.092+(2*-0.11*$\theta$)

$\theta$ = 19.86 radians = 3.16 revolutions