# Homework Help: Calculating velocity needed for uniform circular motion?

1. Oct 4, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Figure 8-32 shows a pendulum of length $L=\frac{5}{4}m$. Its bob (which effectively has all the mass) has speed $v_0$ when the cord makes an angle $\theta_0=40°$ with the vertical. (a) What is the speed of the bob when it is in its lowest position if $v_0=8\frac{m}{s}$? What is the least value that $v_0$ can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight?"

2. Relevant equations
$v^2=v_0^2+2as$
For part (a) $s=L(1-cos\theta)$
For part (b) $s=L(cos\theta)$

$L=\frac{5}{4}m$
$\theta_0=40°$
$v_0=8\frac{m}{s}$

3. The attempt at a solution
(a)
$v^2=v_0^2+2as=v_0^2+2gL(1-cos\theta)$
$v=\sqrt{v_0^2+2gL(1-cos\theta)}=\sqrt{(8\frac{m}{s})^2+2(\frac{98}{10}\frac{m}{s^2})(\frac{5}{4}m)(1-cos(40°))} = \sqrt{64+\frac{49}{2}(1-cos(40°))}\frac{m}{s}=8.35\frac{m}{s}$

(b)
$v^2=v_0^2+2as$
$0=v_0^2-2gL(cos\theta)$
$v_0=\sqrt{2gL(cos\theta)}=\sqrt{(\frac{49}{2})(cos(40°))}\frac{m}{s}=4.33\frac{m}{s}$

And here is what I had trouble with:

I learned in Calc. III that for circular motion to happen, the vector for acceleration must be equal and normal to the vector for velocity. But gravity is always pointing down; it is normal to the velocity vector only at the top and the bottom of the circle. What's more, velocity must be constant throughout the rotation. I started with calculating the work that gravity does in half a rotation.

$W_g=\int_{-L}^{L}(mg)dy=2(mg)L$

Then I attempted to do the same for the circular acceleration:

$W_c=\int_{0}^{L}\frac{mv^2}{r}dy=\int_{-\frac{π}{2}}^{\frac{π}{2}}\int_{0}^{L}(csc\theta)mv^2drd\theta=\int_{-\frac{π}{2}}^{\frac{π}{2}}(Lmv^2)(csc\theta)d\theta=2(Lmv^2)$

I don't get the correct numbers when I equate these two energy values:

$2(mg)L=2(mv^2)L$
$v≠\sqrt{g}$

Can anyone help me? Thanks.

2. Oct 5, 2016

### Eclair_de_XII

Hold on, I did that integration for $csc\theta$ incorrectly...

$W_c=(Lmv^2) \int_{-\frac{π}{2}}^{\frac{π}{2}}csc\theta d\theta=(Lmv^2)(-ln|csc\theta+cot\theta|)=(Lmv^2)(-ln|1+0|+ln|-1|)$

It's still not making sense.

3. Oct 5, 2016

### haruspex

A velocity cannot equal an acceleration.
If its velocity is vt at the top, what is its acceleration there?

4. Oct 5, 2016

### Eclair_de_XII

Is it $\frac{v^2}{r}$? Only at the top, I think, does $g=\frac{v^2}{r}$. At the bottom, $g=-\frac{v^2}{r}$. At the left and right sides, $g$ is tangent to acceleration. Correct me if I am mistaken.

Last edited: Oct 5, 2016
5. Oct 5, 2016

### haruspex

Yes.
Why? there is tension in the string to consider.

6. Oct 5, 2016

### Eclair_de_XII

I think it is because centripetal forces are caused by acceleration of a particle towards its center, while its velocity is tangent to the motion. At $\theta=-\frac{\pi}{2}$, this acceleration points directly upwards, perpendicular to the velocity which points completely in the horizontal direction. Since gravity always points downwards, it's in the negative direction of the acceleration of the particle, which is pointing up at that angle. Since the problem asks for the velocity at which the cord remains taut, gravity must equal acceleration. Gravity and acceleration would cancel each other out, and in turn, the particle would not stray from its path of motion. So the cord, in turn, would remain straight. And I suppose the tension of that cord, in turn, should be equal to acceleration at that point?

Last edited: Oct 5, 2016
7. Oct 5, 2016

### haruspex

At the top of the loop, yes, but not at the bottom. At the bottom, gravity is acting the wrong way to provide centripetal acceleration. The tension in the string at that point must counter gravity, and exceed it by enough to provide the centripetal acceleration.

8. Oct 6, 2016

### Eclair_de_XII

Okay, so one must find the tension necessary to pull the particle one hundred and eighty degrees to the top when gravity is in play. I think I see where work comes into play, here... It's some kind of integration problem, but I just don't know what to put as the limits of integration.

Last edited: Oct 6, 2016
9. Oct 6, 2016

### Eclair_de_XII

This is what I tried:

$W_g=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{L}mgrdrd\theta=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}mgL^2d\theta=\frac{\pi}{2}(mgL^2)$
$W_c=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{L}mv^2drd\theta=mv^2(\pi)(L)$

$\frac{1}{2}(m\pi)(g)(L^2)=(m\pi)(v^2)(L)$
$v^2=\frac{1}{2}Lg$
$v=\sqrt{\frac{1}{2}Lg}=2.47\frac{m}{s}$

10. Oct 7, 2016

### haruspex

It obviously must be more than for case b).
You do not need any integration. You figured out in post #4 what the velocity must be at the top. Use energy conservation to figure out the initial KE.

11. Oct 7, 2016

### Eclair_de_XII

Hm... I tried rearranging things based on the velocity at the top: $\frac{v^2}{r}=g$, which, when solved for $v^2$, becomes $v^2=gr$. And then I multiplied both sides by $\frac{1}{2}m$ to yield kinetic energy. Is this right? $E_k=\frac{1}{2}mgr$.

12. Oct 7, 2016

### haruspex

Yes. So what is the initial KE?

13. Oct 7, 2016

### Eclair_de_XII

Do you mean at the bottom of the circle?

14. Oct 7, 2016

### haruspex

No, at the initial position for part c:

15. Oct 7, 2016

### Eclair_de_XII

Let me see if I got this right...

$E_f=E_k+(mg)(r)(1+cos\theta)$

16. Oct 7, 2016

### haruspex

Depends what you mean by Ef and Ek.

17. Oct 7, 2016

### Eclair_de_XII

$E_k$ is kinetic energy at the very top of the circle and $E_f$ is energy at a point on the circle depending on the position of the particle on the path of motion relative to the center; its angle relative to the center, in other words. If it's at the bottom, $\theta=0$; at the top, $\theta=\pi$.

Last edited: Oct 7, 2016
18. Oct 7, 2016

### haruspex

Ok, but one more thing I should have asked: are you taking g to be a positive quantity or a negative one? Before answering, consider whether the final KE will be more or less than the initial KE, and what your equation says about that.

19. Oct 7, 2016

### Eclair_de_XII

Positive. Gravity will only be adding to $E_k$, I believe.

$E_f=\frac{1}{2}mv^2$
$\frac{1}{2}mv^2=\frac{1}{2}(mgr)+(mgr)(1+cos\theta)$
$v^2=(gr)+2(gr)(1+cos\theta)=(gr)(3+2cos\theta)$
$v=\sqrt{(gr)(3+2cos\theta)}=\sqrt{(\frac{98}{10}\frac{m}{s^2})(\frac{5}{4}m)(3+2cos40°)}=\frac{7}{2}\sqrt{3+2cos40°}=7.45\frac{m}{s}$

Thank you very much. I don't believe I could have done this without your help.

20. Oct 7, 2016

Ok!