Rotational momentum and energy?

Angular momentum is conserved.

L = I * ω

For a solid disk, I = ½ * m * r^2 = ½ * 250 * 2.5^2 =781.25

As it rotates one time, it rotates an angle of 2 π radians

ω = 2 π ÷ 5 = π * 0.4

Li = 781.25 * π *0.4 = 312.5* π

For the boy, I = m * r^2 = 12 * 2.5^2 = 75

Total I = 781.25 + 75 – 856.25

Lf – 856.25 * ω

856.25 * ω = 312.545 * π

ω = 312.5 * π ÷ 856.25

This is approximately 1.154 rad/s. I hope this is helpful for yiu.

let us start at the beginning. What does it mean to place the boy gently on the merry go round.

Usually it means speeding the boy up so that there is no shock loading or collision with the device. This would naturally change the answers.

Conservation of angular momentum indicates initial angular momentum = final angular momentum

i1 w1 = i2 w2

1/2 M r^2 w1 = (1/2 M r^2 + m r^2) w2

multiply BS * 2/r^2

M w1 = (M+2m) w2

w2 = M/(M+2m) w1

= 250/262 * w1 = 250/262 * (2*pi()/5)

I notice that you ALSO asked about energy.

As the energy is 1/2 I w^2

the energy is LESS after placing the boy on the merry go round.

1/2 i2 w2^2

= 1/2 ( 1/2 M r^2 + m r^2 ) * w2^2

= 1/2 *(1/2 Mr^2 + m r^2 )* (M/(M+2m) ) ^2 *w1^2

= 1/4 r^2 (M+2m) * (M/(M+2m)^2 * w1^2

=1/4 * r^2 * (M/(M+2m) ) * w1^2

This is less than

1/2 i1 w1^2

= 1/2 * (1/2 M r^2) w1^2

= 1/4 M r^2 w1^2

Energy is lost by placing the boy on the merry go round.

For you to consider..... Where did it go?

,,,

angular momentum is conserved

omega initial = 1rev/5.00s * 2 pi rad/rev = 1.2566 rad/s

moment of inertia of particle (the boy)

I_boy = m r^2 = 12.0 * 2.5^2 = 75 kg*m^2

moment of disk

I_disk = 1/2 m r^2 = 1/2 (250) (2.5^2) = 781.25 kg*m^2

conservation equation

I omega_i = ( I-disk + I_boy) omega_f

1/2 (250) (2.5^2) * 1.2566 = ( 781.25 + 75) omega-f

856.25 omega_f = 981.72

omega_f = 1.1465 rad/s

round appropriately

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Let I = merry go round moment of Inertia = (1/2)MR^2 M = mass of merry go round and R = radius

The angular momentum initially is: L0 = I*w0 = (1/2)MR^2*(2*pi/5) where w0 = one rev/5 seconds = 2*pi rad/5 sec

THe child adds a moment of inertial Ic = mR^2 but since there are no external torques, angular momentum is conserved:

L0 = L

(1/2)MR^2 *w0 = [ (1/2)M + m]*R^2*w ---> w = {(1/2)MR^2/{[ (1/2)M + m]*R^2} = M/{2*[(1/2)M +m)]} * w0

PLug in for M, m and w0