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In the figure below we have that the gear $A$ is has an fixed axis and has the radius of $R_a = 0,1m$ and, in the instant represented in the figure spins clock-wise with angular velocity $\omega_a = 4\text{rad}/s$. The gear $C$ has radius $R_c = 0,1m$ and gear $D$ has radius $R_d =0,2m$ and they has been firmly weld each other. An arm of length $0,2m$ connects the rotation axis of $A$ and $C$ and $D$. The $B$ gear spins around $A$ but without any connection with $A$ and has radius $R_b = 0,4m$. In the instant presented by the figure $B$ spins in the counter clock-wise and has angular velocity of $\omega_b = 1\text{rad}/s$.

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The question I have is

What is the velocity of the center of gears $C$ and $D$ expressed in $m/s$?

My attempt: I've tried to derive the velocity of the point $v_P$ where $P$ is the point of connection between $A$ and $C$ and I've tried to encounter the angular velocity of $C$ and $D$ using that for a point $Q$ of $B$ must have the velocity equal

$$v_q = \omega_a(L/2) = 0,4m/s$$

Witch leads us to the angular velocity

$$\omega_c = \omega_d = v_q/L$$

But still I couldn't find the velocity using that. Someone have a hint?

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first, I'm not sure that the figure is labelled correctly: it looks like D is colored orange and C is colored yellow. If this is the case, I would assume that $R_C = 0.2m,\; R_D = 0.1m$. After changing those numbers, I found 2 equations on wikipedia: $$N_{s}\omega_{s} + N_{p}\omega_{p} - \left( N_{s} + N_{p} \right) \omega_{c}= 0$$ $$N_{r}\omega_{r} + N_{p}\omega_{p} - \left( N_{r} - N_{p} \right) \omega_{c}= 0$$ Where $\omega$ is speed and N is the number of teeth on each gear. The subscripts are: $s$ for sun (A), $p$ for planet (C,D), $r$ for ring (B), and $c$ for carriage (bar). I'm confident that radius is an acceptable substitute for number of teeth in this case. Hope this helps.

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  • $\begingroup$ I can't use number of teeth to answer that question because I do not have this information. $\endgroup$ – R.W Mar 22 '17 at 13:18
  • $\begingroup$ right, but the system of equations can be scaled by any real number and still give the same answer, right? We could assume that the number of teeth is proportional to the circumference of the gear: $$2 \pi R = N $$ then the first equation is: $$2\pi \left( R_{s}\omega_{s} + R_{p}\omega_{p} -(R_{s} + R_{p})\omega_{c} \right) =0$$ $\endgroup$ – EMiller Mar 22 '17 at 14:13
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In solving many kinematic problems for gears, there is a key rule to key in mind: When two gears are meshing, the tangential velocities of both gears at the point of meshing are equal.

(Also, as said in EMiller's answer, gear C seems to be the smaller orange gear, and gear D the larger yellow gear, contrary to the diagram's labelling).

This rule will help provide the equations we need to solve this problem. Note that there are 2 pairs of meshing gears (A meshes with C and B meshes with D). The points of meshing for each pair are $P$ and $Q$ respectively. Therefore, this gives us 2 equations to work with. It is then a matter of relating the tangential velocities at the point of meshing with the velocity of the centre of gear C and D. I will define the following terms:

Let $v_{a,p}, v_{c,p}$ be the tangential velocities of gears A and C, respectively, at point P.

Let $v_{b,q}, v_{d,q}$ be the tangential velocities of gears A and C, respectively, at point P.

Let $V$ be the velocity of the centre of gear C and D.

(The following diagrams represents gears A and C, and then B and D, as circles that are in pure rolling with one another:)

enter image description here

enter image description here

According to the rule above, we get the following two equations:

$$v_{a,p}=v_{c,p} \qquad (Eq.1)$$ $$v_{b,q}=v_{d,q} \qquad (Eq.2)$$

$v_{a,p}$ and $v_{b,q}$ are simple enough to determine since the centres of gears A and B don't move:

$$v_{a,p}=\omega_a R_a = 0.4\text{m/s} \qquad v_{b,q}=\omega_b R_b = 0.4\text{m/s}$$

Using $Eq.1$ and $Eq.2$, we now know the values for $v_{c,p}$ and $v_{d,q}$. Now, all that remains is to determine the velocity $V$ using the values of $v_{c,p}$ and $v_{d,q}$.

enter image description here

Note that since gears C and D are welded together, they together form a single rigid body. In general, if you know the velocities at two different points on the body, it is possible to determine the velocity at the centre of the body. There are a few ways of determining this, but it may help to determine to centre of rotation and then the angular velocity of gear C/D in order to find the centre velocity.

Hope that helps! :)

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I think I've solved.


This is not an simple exercise.

Solving the problem we must first find the angular velocity of the gears $C$ and $D$. Note that both of them have the same angular velocity because they are strongly fixed one another. For that we use the contact point of $B$ and $D$ that will be (if denoting $L = 0,2m$).

$$v_{\text{Contact point between B and D}} = \omega_D L = \omega_B (2L)$$

Then we have that $\omega_D = \omega_C = 2\text{rad}/s$.

Now we analise the gears sistem $A$ and $C$. In this sistem we need to calculate the center of rotation, wich obviously is in the center of $A$. With that we calculate the velocity of the rotation center of the gears $C$ and $D$.

$$v_{\text{Center of C of D}} = \omega_{\text{Bar}}L$$

We can find the value of $\omega_{\text{Bar}}$ using the contact point

$$v_{\text{Contact point between A and C}} = \omega_{\text{Bar}}\frac{3L}{2} = \omega_C\frac{L}{2} \implies \omega_{\text{Bar}} = \frac{\omega_C}{3}$$

With that we will have

$$v_{\text{Center of C and D}} = \frac{\omega_C L}{3} = \frac{0,4}{3}m/s \approx 0,13m/s$$

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