Question

A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A roundabout of mass 120kg has a radius of 1.0m is initially at rest. A child of mass 43kg is running toward the edge of the roundabout (meaning, running on a path tangent to the edge) at 2.7 m/s and jumps on. Once she jumps on the roundabout, they move together as a single object. Assume the roundabout is a uniform disk. 1. What is the magnitude of her angular momentum (with respect to the center of the roundabout) just before she jumps? 2. What is the angular speed of the roundabout after the jump? 3. Does the overall kinetic energy of the system increase, decrease, or remain constant? If you say it changed, explain what caused a change in energy.

Answer 1

Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

$L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s$

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

$L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s$

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

$K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J$

Final kinetic energy

$K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J$

The overall kinetic energy decreases as can be seen. This loss is converted to heat.