A playground merry-go-round is at rest, pivoted about a friend's axis. A child of mass m runs on the ground along the path tangential to the rim with speed v and jumps on-to-merry go round. If R is the Radius of a merry-go-round and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is:
Answer
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Hint: As the direction of motion of the girl sitting on the merry-go-round is constantly varying, even though the speed is being constant, the velocity at each instant is found to be varying. This vector quantity varying due to the change in direction of motion makes the body have acceleration. Therefore the girl sitting in a merry-go-round is found to be moving with acceleration.
Complete step by step answer:
A playground merry-go-round is at rest, upon a frictionless axis that is there is no external torque acting on the system, angular momentum is conserved.
\[ \Rightarrow \;{{\rm{L}}_i}\; = \;{{\rm{L}}_f}\;\]
$ \Rightarrow \;{\rm{mvR}}\; + \,0 = \left( {{\rm{I + m}}{{\rm{R}}^2}} \right)\;\omega $
Here ${\rm{mvR}}$ is the angular momentum.
If an object is with a massive moment of inertia ${\rm{I}}$, such as Earth, it has a very large angular momentum. An object with large angular velocity $\omega $, such as a centrifuge, also has a rather large angular momentum.
The child's momentum to the center of the merry-go-round and angular momentum of the merry-go-round is zero initially since it is at rest.
$ \Rightarrow \omega \,{\rm{ = }}\dfrac{{{\rm{mvR}}}}{{\left( {{\rm{I + m}}{{\rm{R}}^{\rm{2}}}} \right)}}$
Where,
M= Mass
V= Velocity
R= Radius
I= moment of inertia
Note: Angular momentum- A rigid object's angular momentum is defined as the product of both the moment of inertia and angular velocity.
Angular Momentum = (moment of inertia) (angular velocity)
i) Angular momentum of the merry-go-round is zero initially since it is at rest.
ii) If R is the Radius of the merry-go-round and I is its moment of inertia, then the angular velocity of the merry-go-round and child is zero (pivoted).
Complete step by step answer:
A playground merry-go-round is at rest, upon a frictionless axis that is there is no external torque acting on the system, angular momentum is conserved.
\[ \Rightarrow \;{{\rm{L}}_i}\; = \;{{\rm{L}}_f}\;\]
$ \Rightarrow \;{\rm{mvR}}\; + \,0 = \left( {{\rm{I + m}}{{\rm{R}}^2}} \right)\;\omega $
Here ${\rm{mvR}}$ is the angular momentum.
If an object is with a massive moment of inertia ${\rm{I}}$, such as Earth, it has a very large angular momentum. An object with large angular velocity $\omega $, such as a centrifuge, also has a rather large angular momentum.
The child's momentum to the center of the merry-go-round and angular momentum of the merry-go-round is zero initially since it is at rest.
$ \Rightarrow \omega \,{\rm{ = }}\dfrac{{{\rm{mvR}}}}{{\left( {{\rm{I + m}}{{\rm{R}}^{\rm{2}}}} \right)}}$
Where,
M= Mass
V= Velocity
R= Radius
I= moment of inertia
Note: Angular momentum- A rigid object's angular momentum is defined as the product of both the moment of inertia and angular velocity.
Angular Momentum = (moment of inertia) (angular velocity)
i) Angular momentum of the merry-go-round is zero initially since it is at rest.
ii) If R is the Radius of the merry-go-round and I is its moment of inertia, then the angular velocity of the merry-go-round and child is zero (pivoted).
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