I need to find the sum of energy in this device :
A green support turns clockwise at w. A motor is fixed on that support. The stator of this motor turns clockwise at w (it is on the support). The rotor of the motor turns clockwise at 2w (lab reference). On the rotor there is a big pulley (P1) with a radius of 3R. That pulley P1 drives a smaller pulley P2 with a belt. The pulley P2 has a radius R. The small pulley P2 is braking on the support at the radius R. There is friction so an energy from heating. The force of the friction is a constant F.
I drawn a figure:
imagizer.imageshack.us/a/img661/4682/jIXu8r.png
The motor gives the forces F1/F2 to the big pulley. The motor gives the forces F3/F4 to the support. Axes of pulleys give forces F6/F8 to the support. The brake gives F8 to the support and gives F9 to the small pulley.
For me, the motor needs the energy -3FRwt because the rotor turns at 2w but the stator turns at w. There is no torque on the support: torque from F3/F4 is cancelled wih F6/F8. The brake gives the heating 5FRwt. Because the small pulley P2 turns at 6w (lab reference).so on the support the small pulley turns at 6w-w=5w, so for me the energy from heating is 5FRwt. The sum for me is 2FRwt, Where I'm wrong ?
To find the sum of energy in the device, let's break down the different components and the energy associated with each.
1. Motor: The motor is rotating at a speed of w (clockwise) on the support. The stator of the motor is turning at the same speed, w. The rotor, on the other hand, is turning clockwise at 2w relative to the lab reference. The motor requires energy input to drive the rotor, and the energy required can be calculated as -3FRwt. This negative sign indicates the energy being supplied by an external source to the motor.
2. Friction in the small pulley: The small pulley, P2, is braking on the support at a radius of R. This creates friction, which leads to energy loss in the form of heat. The force of friction is constant, with a value of F. As the pulley turns at 6w according to the lab reference, but only moves at 5w on the support (due to the stator turning at w), the energy loss due to heating can be calculated as 5FRwt. This positive sign indicates energy being dissipated as heat.
Now, let's calculate the sum of energies:
Sum of energies = Energy input - Energy loss due to heating
= (-3FRwt) - (5FRwt)
= -8FRwt
Therefore, the sum of energies in the device is -8FRwt.
It seems that you may have made an error in your calculation by not accounting for the negative sign associated with the energy input from the motor. The correct sum of energies is -8FRwt, not 2FRwt.