0.90 m 3. A child of mass 29 kg runs and jumps onto a playground carousel that is initially at rest. The carousel can be approximated as a solid disk of mass M 410 kg and radius R-1.4 m, with moment of inertia l- 1R2. Before jumping on the carousel, the child moves along a line that is a perpendicular distance of 0.90 m from the center of the carousel (as shown) with a constant speed of 11 m/s. The child jumps onto the edge of the carousel as shown (a) Consider the system consisting of the child and the carousel, and consider the initial situation to be the just before the child lands on the carousel; the final situation is just after the child lands on the carousel. Is kinetic energy conserved? Explain (b) Is linear momentum conserved? Explain (c) Is angular momentum conserved? Explain (d) Find the final rotational speed of the carousel with the child on it Solution Q3. mass of child =m=29 kg mass of the solid disk=M=410 kg radius R=1.4 m moment of inertia=I=(1/2)*M*R^2 =401.8 kg.m^2 initial speed of the child=v=11 m/s distance from the centre=r=0.9 m conserving angular momentum: initial angular momentum of the disk +initial angular momentum of the child=final angular momentum of the disk + child system let final speed of the disk + child system be v1 m/s. their angular speed=linear speed/radius =v1/R then conservation of angular momentum principle states: 0moment of inertia of the disk*initial angular speed+mass of the child*speed of the child*perpendicular distance=total moment of inertia of the disk and child system*angular speed ==>0+m*v*r=(I+m*R^2)*(v1/R) ==>m*v*r*R=(I+m*R^2)*v1 substituting the values for all the symbols, we get v1=0.8764 m/s initial kinetic energy=0.5*m*v^2=1754.5 J final kinetic energy=0.5*(I+m*R^2)*(v1/R)^2=89.86 J part a: as initial kinetic energy is not equal to final kinetic energy, kinetic energy is not conserved. part b: linear momentum is conserved as there is no external force. part c: angular momentum is conserved, as there is no external torque applied on the system part d: final rotational speed=v1/R =0.626 rad/s .

1. 0.90 m 3. A child of mass 29 kg runs and jumps onto a playground carousel that is initially at
rest. The carousel can be approximated as a solid disk of mass M 410 kg and radius R-1.4 m,
with moment of inertia l- 1R2. Before jumping on the carousel, the child moves along a line that
is a perpendicular distance of 0.90 m from the center of the carousel (as shown) with a constant
speed of 11 m/s. The child jumps onto the edge of the carousel as shown (a) Consider the system
consisting of the child and the carousel, and consider the initial situation to be the just before the
child lands on the carousel; the final situation is just after the child lands on the carousel. Is
kinetic energy conserved? Explain (b) Is linear momentum conserved? Explain (c) Is angular
momentum conserved? Explain (d) Find the final rotational speed of the carousel with the child
on it
Solution
Q3.
mass of child =m=29 kg
mass of the solid disk=M=410 kg
radius R=1.4 m
moment of inertia=I=(1/2)*M*R^2
=401.8 kg.m^2
initial speed of the child=v=11 m/s
distance from the centre=r=0.9 m
conserving angular momentum:
initial angular momentum of the disk +initial angular momentum of the child=final angular
momentum of the disk + child system
let final speed of the disk + child system be v1 m/s.

2. their angular speed=linear speed/radius
=v1/R
then conservation of angular momentum principle states:
0moment of inertia of the disk*initial angular speed+mass of the child*speed of the
child*perpendicular distance=total moment of inertia of the disk and child system*angular speed
==>0+m*v*r=(I+m*R^2)*(v1/R)
==>m*v*r*R=(I+m*R^2)*v1
substituting the values for all the symbols, we get
v1=0.8764 m/s
initial kinetic energy=0.5*m*v^2=1754.5 J
final kinetic energy=0.5*(I+m*R^2)*(v1/R)^2=89.86 J
part a:
as initial kinetic energy is not equal to final kinetic energy, kinetic energy is not conserved.
part b:
linear momentum is conserved as there is no external force.
part c:
angular momentum is conserved, as there is no external torque applied on the system
part d:
final rotational speed=v1/R
=0.626 rad/s