# Instant centre of rotation for two connected gears

The two gears are have the angular velocities $\omega_1$ and $\omega_2$ respectively with respect to $Oxyz$. The task is to determine the angular velocity $\boldsymbol{\omega}$ of the arm $OA$.

Denote the point of contact between the two gears with $C$. At that point, the smaller gear gives the larger gear a velocity of $\textbf{v}_C = \omega_2 r \, \boldsymbol{e}_y$. Velocities of two points in a rigid body relate by $\textbf{v}_1 = \textbf{v}_2 + \boldsymbol{\omega} \times \textbf{r}_{21}$, hence

$$\textbf{v}_O = \omega_2 r \, \textbf{e}_y + \omega_1 \textbf{e}_z \times (-R) \, \textbf{e}_x = (\omega_2 r - \omega_1 R) \, \textbf{e}_y \ .$$

Similarly, at point $C$, the larger gear gives the smaller gear a velocity of $\textbf{v}_C = \omega_1 R \, \textbf{e}_y$. Hence

$$\textbf{v}_A = \omega_1 R \, \textbf{e}_y - \omega_2 \, \textbf{e}_z \times r \, \textbf{e}_x = ( \omega_1 R - \omega_2 r) \, \textbf{e}_y \ .$$

Analyzing the velocities of point $A$ and $C$ of the arm $OA$, we have

$$\begin{gather} ( \omega_1 R - \omega_2 r) \, \textbf{e}_y = (\omega_2 r - \omega_1 R) \, \textbf{e}_y + \boldsymbol{\omega} \times (R + r) \, \textbf{e}_x \\ \iff \\ 2( \omega_1 R - \omega_2 r) \, \textbf{e}_y = \omega (R +r ) \, \textbf{e}_y \end{gather}$$

Hence $$\omega = \frac {2( \omega_1 R - \omega_2 r) }{R+r} \ .$$

It turns out that this in fact is incorrect. Why exactly, I am not sure. According to the key, the point of contact $C$ is an instant centre of rotation for each gear i.e. $\textbf{v}_C = \textbf{0}$ for each gear. This will produce

$$\omega = \frac {( \omega_1 R - \omega_2 r) }{R+r} \ ,$$

i.e. my result was twice as large.

How is it that the gears do not influence each other at the point $C$? In similiar problems that I have done, the velocity of the point of contact of gears is influenced by the other turning gears. But in this problem all of sudden there is no such influence? What is at fault in my understanding of instant centres of rotation?

• Is O fixed in space and A orbiting around? Commented Apr 21, 2015 at 3:34

Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car.

As far as I understand, confusion point is that first you calculate VO in respect of point A that is moving itself. In the next line you calculate VA in respect of point O that is moving too, right in the opposite direction to point A. That's why the result is doubled.

I would first assume the large gear stopped (it means take a look in reference frame of the larger gear). Here point C is not moving, so point A has velocity VA = - w2 * r (negative because A goes downwards). Then allow the larger gear to rotate (it means take a look in reference frame Oxyz). Now we have to add upwards (positive) velocity VC to VA, so will get VA = - w2 * r + VC = - w2 * r + w1 * R. Finally take angular velocity of the arm as velocity of point A in respect of O, divided by length of the arm, that is (R+r). So obtain w = (w1*R - w2*r) / (R+r).

Apparently there are many ways to deals with this problem. For me it seems just easier to take a look at relative reference frame first.

• Point C is a relative instant center of rotation. In an inertial frame $Oxyz$ point C has tangential velocity. Commented Nov 9, 2016 at 21:18
• Agree, @ja72. That's why we have to add the term ( w1 * R ) that is relative speed VC to change relative frame onto Oxyz. This gives us final solution. Commented Nov 14, 2016 at 19:42

I would approach this problem by "thinking" that the gear $A$ is like on a treadmill. So the gear $O$ is like moving "backward" while the gear $A$ is trying to move forward. You can easily compute the velocity of the two contact points and see what happens.

• Could please show me how exactly you had in mind? Thanks Commented Feb 4, 2015 at 14:55
• If you linearise the problem then gear O is shifting the ground, say leftward, with speed $\omega_1R$, while the gear A moves rightward on it with speed $\omega_2r$. The relative speed will give you the motion of the arm. I think you can now take it from here. Commented Feb 5, 2015 at 9:18
• What do you mean by linearising? I don't see what is non-linear about the problem. The velocity of the contact points is apparently zero according to key, are you suggesting they are non-zero? Do you arrive at the correct answer, i.e. the correct angular velocity of the arm, with your approach? Much appreciated if you could elaborately outline what you mean! Commented Feb 5, 2015 at 14:05

This is a problem that involves only calculation of velocities from other velocities, no influence (forces) needs to be considered.

Imagine the system as it appears in the inertial system where the point of contact of the two wheels is at rest.

Since the contact point is at rest, the mass points of both wheels that touch each other are at rest. Since any planar motion of rigid body is a rotation in this plane, this means that the contact point is a center of rotation of both wheels.

The angular velocities are the same as in the original inertial frame, $\omega_1$ , $\omega_2$.

The velocity of the center 1 is then $\omega_1 R$, velocity of the center 2 is $\omega_2r$. The endpoints of the arm have the same velocities.

The arm is a also a rigid body in planar motion, so it rotates in this plane. It rotates about some point $P$ somewhere on the infinite axis defined by the two wheel centers. Let use denote distance of this point $P$ from the point $A$ by $X$.

The angular velocity of the arm around $P$ has to be such that the endpoints have above velocities. This leads to equations

$$\omega X =\omega_1 R,$$

$$\omega (X-R-r) = \omega_2 r.$$

The $\omega$ that solves these two equations is indeed

$$\frac{\omega_1 R - \omega_2 r}{R+r}.$$

Position of the point $P$ - coordinate $X$ - can also be calculated.

Why is your answer incorrect? Because at first you're calculating velocity of $C$ assuming point $A$ is at rest. Then the velocity you find for point $O$ is its velocity relative to the point $A$. Similarly, when you're calculating the velocity of point $A$ what you find is actually its velocity relative to $O$. They are the same thing, only with a sign difference, so when you add them together you get twice the correct answer.

Center or rotation isn't located where those two gears make contact, it's in a point closer to the smaller gear's center

• The definition of the instant center of rotation is at the point of contact (cf. this Wikipedia entry). Commented Nov 14, 2015 at 17:12