# Angular Acceleration

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1. Oct 8, 2016

### Jonski

1. The problem statement, all variables and given/known data

In the mechanism shown below, the angular velocity of link 2 is 17 rad/s CW, constant. Find the angular velocity (magnitude and direction) of link 3. The distance between A and C is 109 mm, the length of link 3 is 125 mm, and θ= 111°. Note: B is a pin-slider

In the mechanism shown above, find the relative velocity (magnitude) of point B with regards to link 2.

In the mechanism shown above, find the direction of the relative velocity of point B with regards to link 2. Take the right-hand direction as a reference axis and CCW positive.

In the mechanism shown above, determine the angular acceleration (magnitude and direction) of link 3.

2. Relevant equations
a3 = a3t + a3n + a rel + a cor
a3t = αr
a3n = ω^2r
a cor = 2ωv
3. The attempt at a solution
So I have done the first 3 parts of the question correctly getting
0.800m/s = Vb
-291deg

I am not sure how to get the acceleration though. I'm not sure how to find the tangential component and the relative acceleration.
Thanks

2. Oct 8, 2016

### Staff: Mentor

You have the magnitude and the velocity of point B. If BC is of constant length, what is the component of the velocity of point B along the link BC? What does this tell you about the component of the velocity of point B perpendicular to the link BC?

3. Oct 8, 2016

### Jonski

Does it have something to do with the fact that since the angular velocity of bar AB is constant the tangential acceleration of bar AB is 0.

4. Oct 8, 2016

### Staff: Mentor

No. Please try to answer my questions. If you have trouble with the first question, I'll tell you the answer.

5. Oct 8, 2016

### Jonski

Wouldn't the velocity perpendicular to BC be equal to:
Vb = Vb' + Vb/b'
Vb = 17(0.0335) + 0.8
Vb = 1.37 m/s

6. Oct 9, 2016

### Staff: Mentor

I originally misinterpreted this question, so I'm going to start over by first checking your answers. Let $\alpha$ be the angle ABC and $\phi$ be the angle ACB. Using the law of sines, $$\frac{\sin{\alpha}}{109}=\frac{\sin{\theta}}{125}$$
So $\alpha=54.5$ degrees. That means that $\phi = 180 -111-54.5=14.5$ degrees.
Let r equal the distance of point B from point A. Again using the law of sines:
$$\frac{\sin 14.5}{r}=\frac{\sin{\theta}}{125}$$Solving for r yields: $r=33.53$ mm.

The law of cosines tells us that: $$r^2+109^2-2(109)r\cos{\theta}=125^2$$
If we differentiate this equation with respect to time, we obtain: $$2r\frac{dr}{dt}-2(109)\cos{\theta}\frac{dr}{dt}+2(109)r\sin{\theta}\frac{d\theta}{dt}=0$$
Solving for dr/dt yields:
$$\frac{dr}{dt}=-\frac{109r\sin{\theta}}{r-109\cos{\theta}}\frac{d\theta}{dt}=\frac{109r\sin{\theta}}{r-109\cos{\theta}}\omega\tag{1}$$
Substituting our values for r and $\theta$ in this equation yields:$$\frac{dr}{dt}=808\ mm/s$$This is the velocity of point B relative to link 2, and is very close to your answer.

The law of sines tells us that $$\frac{\sin{\phi}}{r}=\frac{\sin{\theta}}{125}$$
Differentiating this equation with respect to time yields:
$$\cos{\phi}\frac{d\phi}{dt}=\frac{\sin{\theta}}{125}\frac{dr}{dt}-\frac{r\cos{\theta}}{125}\omega\tag{2}$$
From this, I get that the angular velocity of link 3 is 7.94 radians per second CW. This is also close to your answer, but not quite.

For the direction of the velocity of B relative to link 2, I get that it is along link 2.

To get the angular acceleration of link 3, I would differentiate Eqns. 1 and 2 with respect to t and combine the results, taking into account that $d\theta /dt$ is constant.