10 Points A conducting solid cylinder of mass m length Lresi

(10 Points) A conducting solid cylinder of mass \"m\", length \"L\",resistance \"R and radius\"r is placed on the rails at t-0 as shown. The rails are inclined 30 degrees from the horizontal and the cylinder starts at an initial position \"x,from the upper end. A battery of voltage-V\" is connected to the upper end of the rails and the lower end is open. The rails have no resistance. There is a constant magnetic field \"B projected vertically upward through the rails. When the cylinder is placed on the rails the circuit is closed causing a current in the circuit. The static friction between the cylinder and rails is large enough so that no slippage occurs as the cylinder rolls down. Given [m, L. R,r. V, B, C,] Determine: a·The magnetic flux through the circuit as a function of position. b. The induced EMF as a function of velocity c. The current in the circuit as a function of velocity d. The magnetic force on the cylinder as a function of velocity e. The velocity of the cylinder center of mass as a function of time f. The velocity of the cylinder after a long time (terminal velocity). g. The kinetic energy (translational and rotational) as a function of time h. The power output of the battery as a function of time i. The power lost to heat as a function of time. j. Prove that the power given to the circuit from the rate of change in gravitational potential energy added to the power produced by the battery is equal to the rate of change in the kinetic energy added to the power lost to heat (10 Points) After a long time, the cylinder runs off the rails and onto another set of rails. The new set of rails has no battery connected to it and a wire with no resistance ties the far end of the rails together. Reset the position and time to zero at this point. Determine k·The velocity of the cylinder as a function of time. 1. The distance the cylinder rolls before coming to rest. m. The current as a function of time. n. Prove that the original kinetic energy is equal to the heat dissipated by the resistor (10 Points) Repeat the second part where the cylinder runs off the rails and onto a horizontal set of rails. Now the new set of rails has instead a capacitor \"C\" connected across their far ends. Again the static friction is large enough so that no slippage occurs. Reset the position and time to zero at this point. Determine: . o. The velocity of the cylinder as a function of time. p. Plot the velocity as a function of time q. What is the terminal velocity of the cylinder? r. The current as a function of time. s. Prove that on this horizontal section, the change in kinetic energy is equal to the energy of the capacitor plus the energy dissipated by the resistor Mat Poue Kdhe resi

Solution

Write 2 equations, first applying krichoff\'s law in the loop
BLv - R i - q/C = 0   .........1
Second is of force on conductor due to magnetic field.
2BLi/3 = -m dv/dt   .......2    ( factor of 2/3 is due to rolling)
eliminating i between 1 and 2 we get
BLv + R(3m dv/dt)/2BL - q/C = 0    ......3

Writing i = rate of change of charge on capacitor = dq/dt, second equation may be written as
2 BL dq / 3 = - m dv
initial q =0 , v= vo   ( vo is speed of cylinder as it enters second rail)
Hence 2BL q/3 = m (vo - v)
this gives q as function of v as
q = 3m(vo-v) / 2BL   (That is what you were looking for)
Substitute this in equation 3 and integrate to finally get v as function of time.