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Angular velocity and energy lost due to friction

  • Thread starter mossman
  • Start date
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1. Homework Statement

the problem is: A 3.0 m diameter merry go round with rotational inertia 120kgm^2 is spinning freely at 0.50 rev/s. Four 25kg children sit suddenly on the edge of the merry go round. a) Find th new angular speed, and b) detrmine the total energy lost to friction between the children and merry go round.

2. Homework Equations
conservation of angular momentum = L_initial = L_final
angular momentum = Iomega^2
conservation of rotational mechanical energy = KE_final - KE_initial = 0 + W_nonconservative


3. The Attempt at a Solution

So, r = 1.5m, I_wheel = 120 kgm^2, omega_wheel = 0.5 rev/s, and since when the children jump onto the wheel, we can treat them as a particle of the whole system of the wheel and children, so we take the distance of the children from the center of the wheel, so I_kids=25*4*1.5^2, I_kids = 225kgm^2, and since there is no external torque, angular momentum is conserved, so we use L_initial = L_final, so L_initial = I_wheel*omega_wheel, and L_final = (I_kids+I_wheel)*omega_final, so L_final = (225+120)*omega_final = 120*0.5 = 60kgm^2rad/s, so 345omega_final = 60, omega_final = 60/345 = 0.17 rad/s, and so now we can use the conservation of rotational energy, W_net = W_conservative + W_nonconservative = KE_final - KE_initial = 0 + W_nonconservative, so 345*(0.17)^2/2 - 120*(.5)^2/2 = W_nc = 5 - 15 = -10 kgm^2rev/s, so 10 kgm^2rev/s of energy is lost, if we convert it to radians we get -10kgm^2rev/s*2Pirad/1rev = -20_Pi_kgm^2rad/s
 

Answers and Replies

35
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So, r = 1.5m, I_wheel = 120 kgm^2, omega_wheel = 0.5 rev/s, and since when the children jump onto the wheel, we can treat them as a particle of the whole system of the wheel and children, so we take the distance of the children from the center of the wheel, so I_kids=25*4*1.5^2, I_kids = 225kgm^2, and since there is no external torque, angular momentum is conserved, so we use L_initial = L_final, so L_initial = I_wheel*omega_wheel, and L_final = (I_kids+I_wheel)*omega_final, so L_final = (225+120)*omega_final = 120*0.5 = 60kgm^2rad/s
A 3.0 m diameter merry go round with rotational inertia 120kgm^2 is spinning freely at 0.50 rev/s.
0.5 rev/s (Hz) is the frequency, not the angular velocity. [tex]\omega=2\pi f[/tex].
 
3
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otherwise everything else is correct?
 

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