# Different approaches gives different results: planetary gear problem?

I am having a homework:

"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 ?

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.

• Is the missing torque is the torque to keep gear 1 stationary?. Without it, gear 1 would be moving?
– Dat
Dec 2 '19 at 11:54
• 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. Dec 2 '19 at 12:37
• (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. Dec 2 '19 at 12:43
• 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 $\gamma$
– Dat
Dec 2 '19 at 16:50
• 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? Dec 2 '19 at 17:51