Question

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of ω = 1.53 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 73 kg and velocity v = 4.2 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.1)What is the magnitude of the initial angular momentum of the merry-go-round?kg-m2/sYour submissions:2)What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?kg-m2/sYour submissions:

Answer 1

Answer:

299.88 kgm²/s

499.758 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.63 m

I = Moment of inertia = 196 kgm²

$\omega_i$ = Initial angular velocity = 1.53 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.2 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s$

The initial angular momentum of the merry-go-round is 299.88 kgm²/s

Angular momentum is given by

$L=mvR\\\Rightarrow L=73\times 4.2\times 1.63\\\Rightarrow L=499.758\ kgm^2/s$

The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s