Big Chris takes little Sam to the playground to ride the children\'s frictionles

Question

Big Chris takes little Sam to the playground to ride the children's frictionless merry-go-round. Which can be modeled as a uniform disk of mass = 1000(kg) and diameter = 4.0(m). It rotates about a vertical axis through the center of the disk. The merry-go-round is initially at rest when Big Chris applies a constant tangential force of 500(N) to the rim for 10(s). Calculate the resulting angular velocity, angular momentum and rotational kinetic energy of the merry-go-round

B. Big Chris places little Sam whose mass is 20(Kg) on the rim. Calculate the final angular velocity and rotational kinetic energy, and what fraction of the initial rotational kinetic energy is converted to heat. Little Sam can be modeled as a point mass.

Explanation / Answer

The disk has a mass M = 1000kg and a radius R = 2m, so its moment of inertia is

I = (1/2)MR^2 = (1/2)(1000kg)(2m)^2 = 2000kgm^2

a) The torque = Fr = I*, and

= I/Fr = (2000kgm^2)/(500N*2m) = 2 rad/s^2

After 10s, = t = (2 rad/s^2)(10s) = 20 rad/s,

L = I = (2000kgm^2)*(20 rad/s) = 40,000 kgm^2/s

K = (1/2)I^2 = (1/2)(2000kgm^2)(20rad/s)^2 = 400,000J

b) If you add a mass to the edge, you add a moment of inerta

I = mR^2 = (20kg)(2m)^2 = 80 kgm^2, so the new I' = 2080 kgm^2.

Angular momentum is conserved, so L = L' and

I = I'', so ' = (40,000 kgm^2/s)/(2080 kgm^2) = 19.2 rad/s

K' = (1/2)I'(')^2 = (1/2)(2080kgm^2)(19.2rad/s)^2 = 383,385J

So, the fraction converted to heat is 1 - K'/K = 1 - (383,385J)/(400,000J) = 0.04 = 4%

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