I am having a homework: enter image description here

"A planetary gear system with a fixed gear 1 (radius r1); gear 2 (radius r2) is movable". At begining, the system is stationary. Apply a constant torque M to OA bar. OA bar rotates about O and cause gear 2 moving. OA has weight Q, Gear 2 has weight P. Calculate the angular acceleration of OA bar."

I am doing this homework with two approach and they gives different answers:

Approach 1: Energy method

Let the angular velocity of OA bar is $\omega$

Kinetic energy of OA bar = $\frac{1}{2}\frac{Q}{g}\frac{(r_1+r_2)^2}{3}\omega^2$

Kinetic energy of gear 2 = $\frac{1}{2}(\frac{1}{2}\frac{P}{g}r_2^2)\omega_2^2+ \frac{1}{2}\frac{P}{g}v_A^2$

$\omega_2 = \frac{r_1+r_2}{r_2}\omega$

$v_A = (r_1+r_2)\omega$

Hence, total kinetic energy = $\frac{1}{2}\frac{2Q+9P}{6g}(r_1+r_2)^2\omega^2$ = total work = M$\phi$

Differentiate two side, give angular acceleration $\gamma$= $\frac{6Mg}{(2Q+9P)(r_1+r_2)^2}$

Approach 2: angular momentum method

Angular momentum of OA with respect to point O = $\frac{1}{3}\frac{Q}{g}(r_1+r_2)^2\omega$

Angular momentum of gear 2 with respect to point A = $\frac{1}{2}\frac{P}{g}r_2^2\omega_2$

Angular momentum of gear 2 with respect to point O = Angular momentum of gear 2 with respect to point A + $\frac{P}{g}OAv_A$ = $\frac{1}{2}\frac{P}{g}r_2^2\omega_2 + \frac{P}{g}\omega(r_1+r_2)^2$

Hence, total angular momentum of system with respect to point O = $\frac{1}{3}\frac{Q}{g}(r_1+r_2)^2\omega + \frac{1}{2}\frac{P}{g}r_2^2\omega(r_1+r_2)/r_2 + \frac{P}{g}\omega(r_1+r_2)^2$

Differentiate above term give us: $\gamma(\frac{1}{3}\frac{Q}{g}(r_1+r_2)^2 + \frac{1}{2}\frac{P}{g}r_2^2(r_1+r_2)/r_2 + \frac{P}{g}(r_1+r_2)^2) = M$

Hence $\gamma = \frac{6Mg}{(2Q+9P)(r_1+r_2)^2-3Pr_1(r_1+r_2)}$

The two results are different, what I am missing ?


1 Answer 1


Approach 1 is correct.

You haven't considered all the torques acting on the system (rod + gear 2) in approach 2.

The force corresponding to the missing torque, is responsible for maintaining the pure-rolling motion of gear 2 (constraint: point of contact instantaneously at rest) on the surface of gear 1 at all times. The torque you're missing doesn't do work on the system which is why approach 1 gave the correct answer even though you didn't realize the presence of this torque. I will leave you to figure out this missing torque.

  • $\begingroup$ Is the missing torque is the torque to keep gear 1 stationary?. Without it, gear 1 would be moving? $\endgroup$
    – Dat
    Dec 2, 2019 at 11:54
  • $\begingroup$ No. The missing torque arises from the force of interaction between gear 1 and 2: the force is very similar to the frictional force that is responsible for enforcing the "no-slipping" constraint between the two discs. This missing torque acting on gear 1 can be counter-acted by the torque applied by the ground/pivot to give a zero net torque: That's why gear 1 remains stationary. And this missing torque acting on gear 2 is responsible for the angular acceleration of gear 2 ($\dot{\omega}_2$). To see this, write down $\frac{dL_{CM}}{dt}= \tau_{CM}$ considering the system to be gear 2 alone. $\endgroup$
    – Ajay Mohan
    Dec 2, 2019 at 12:37
  • $\begingroup$ (contd.) CM is the center of mass of gear 2: point A in the figure. Your first step should be to find the force of interaction between gear 1 and gear 2 that's tangential to the two discs. Let me know if you're unable to understand/proceed. $\endgroup$
    – Ajay Mohan
    Dec 2, 2019 at 12:43
  • $\begingroup$ I can only express the force of interation between gear 1 and gear 2 to a function of $\gamma$. Then add momen of this force with respect to O in approach 2, then solve the equation for $\gamma$ $\endgroup$
    – Dat
    Dec 2, 2019 at 16:50
  • $\begingroup$ Yes, it is a function of $\gamma$ since $\dot{\omega}_2$ is a function of $\gamma$. Did the result from approach 2 now match with the result from approach 1? $\endgroup$
    – Ajay Mohan
    Dec 2, 2019 at 17:51

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