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Mechanical Engineering

Sundaram Ramchandran

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  1. The above could possibly be extended to laser cooling in general (applicable to all phases - gaseous , fluid and solid)
  2. I guess the essential relationships in terms of angular velocities still hold, irrespective of the shape of the arm / carrier which would , i suppose , still fulfill its function of transmitting the input angular momentum to the gears attached to it.
  3. An alternative answer would be not to subtract wPC from wR (since the ring gear is not attached to the arm) , in which case, wR / (wS - wPC) = X. That is , (wS - wPC) * X = wR = 0. Since X is non-zero, we would have wS = wPC, That is , when the ring gear is fixed, the angular velocities of the arm / planet carrier and the sun gear are equal and in the same direction. I was somehow not satisfied with this answer. Using the same logic, for question 3, we would get wR / (3500 - (- 1575) = wR / 5075 = X = .042576. Therefore wR = 5075*.042576 = 216.0732 in the same direction as S , anti-clockwise
  4. The book is : "Theory of Machines and Mechanisms - Dicker, Pennock and Shigley" - 3rd Edition, Oxford University Press, 2003 . Further details ( Chapter 10 - Mechanism Trains, equation 10.4 and solved example 10.2 ). I realized that the energy calculations would be considerably more complex involving : - The kinetic energy (KE) of the sun gear (with its moment of inertia etc) - The KE of the Planet Carrier / arm (whose mass / moment of inertia would possibly include those of the planet gears) The above would be the input energy and would be distributed between the KE's of the 3 sets of planet gears (only 1 of which is shown in the diagram) and the KE of the ring gear The KE of the planet gears would involve both translational KE due to the velocity of the center of mass as well as their rotational KE's involving moment of inertia etc. The KE of the ring gear is purely rotational. I did not attempt this because it is considerably more involved !!
  5. If we look at each point along the curve, we could possibly compute the force Mg*sin(theta) - mg where M is the mass of the wheel and m is the mass to be lifted. This integrated over an infinitesimal unit of arc length would give us work done , say W1 . W1 - Iw^2 /2 (I is the moment of inertia of the wheel and w is the angular velocity of the wheel) = Mv^2 / 2 (where v is the velocity of center of mass of wheel). The velocity of the wheel will be the velocity of the rope and the velocity with which mass m is hoisted up since they are connected. Since the wheel does not slip, the angular velocity w and velocity of center of mass v can be linked. Possibly, the force (M*g*sin(theta) - mg) can be connected to the contact point velocity ???
  6. The Kinematic coeff say X) , i.e the ratio (wR-wPC) /(wS - wPC) = (-Ns / N1) * (N2 / N3) * (-N4 / Nr) where (wR - wPC) is the angular velocity of the ring gear relative to wPC , the angular velocity of the planet carrier. Likewise, wS is the angular velocity of the sun gear and wS - wPC the relative angular velocity ( though the ring gear is not fixed to the arm unlike the other gears, we are measuring velocities relative to the planet carrier, hence the subtraction). But since this is stationary , wR is zero. Hence (- wPC / wS - wPC) = X. . Therefore, wS*X - wPCX = -wPC . Therefore, -wS*X = (1-X)*wPC Therefore, wPC / wS = - X / (1 - X). In our case, the kinematic coeff X = (-16 / 29) * (17 / 35) * (-17/107) = 4624 / 108605 = .042576 (app) . Therefore, in our case, wPC / wS = -.042576 / (1-.042576) = - .04447 which is in the opposite direction. In the 3rd problem, if we take clockwise rotations as positive, then wS - wPC = 3500 - (-1575) = 5075 and the relative velocity of the ring gear , wR - wPC = 5075*X = 216.07 (app) . Here, i was confused whether to add the velocity of carrier arm to the relative angular velocity obtained above. But going by the above argument, to be consistent one would have to add back the arm velocity. Hence, one would have to subtract 1575 , since the angular velocity of the planet carrier arm is in the anti-clockwise direction. Therefore, angular velocity of the ring = 216.07 - 1575 = - 1358.925, i.e 1358.925 in the anti-clockwise direction
  7. One innovative aspect relates to the possibility of using solid state laser cooling to accelerate by leaps and bounds what seems to be the main bottleneck in metallurgical (or any material ) formation / fabrication (starting from extraction) processes, i.e the thermodynamic aspects of heating / cooling (which can now be carried out using precisely controlled laser power pulses / waves
  8. I have the following scenario in mind : - A series (few in number)of electrically driven rotary gears (pinions) in a rack and pinion system are in contact with a rack. Would it be fair to assume that the displacement of the rack would be a superposition of the motions of the rotors (and hence the driving electrical waveforms ) ?? - If only some of the rotors are electrically driven, would the above still be the case ? For example, when only one of the rotors is driven, would the displacement of the rack still track the driving waveform(s) or would the passive rotors significantly affect the transfer of energy / momentum ? (Assume that the rotors are lightweight , small and have low inertia and there is enough lubrication to avoid wear and tear etc ). Thanks, Sundar
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