# Time for bead to slide along the chord of a vertical circle

I ran into this problem in my mechanics homework.

Here's my go at it. I hit a wall at the end and I just don't know what to do.

assuming this circle

Please note that $$\alpha \neq 90$$ degrees. It's just faulty sketching. Sorry.

$$\because Arc Length (L) = 2rSin(\frac{\theta}{2})$$

$$\therefore S_1 = 2rSin(\frac{\theta}{2}), S_2 = 2rSin(\frac{\alpha}{2})$$

$$\because S = V_i + \frac{1}{2}at^2$$

$$\therefore S_1 = 0 + \frac{1}{2}a(t_1)^2, S_2 = 0 + \frac{1}{2}a(t_2)^2$$

$$\therefore 2rSin(\frac{\theta}{2}) = \frac{1}{2}a(t_1)^2, 2rSin(\frac{\alpha}{2}) = \frac{1}{2}a(t_2)^2$$

$$(t_1)^2 = \frac{4rSin(\frac{\theta}{2})}{a}$$, $$(t_2)^2 = \frac{4rSin(\frac{\alpha}{2})}{a}$$

$$\therefore \frac{(t_1)^2}{(t_2)^2} = \frac{\frac{4rSin(\frac{\theta}{2})}{a}}{\frac{4rSin(\frac{\alpha}{2})}{a}}$$

$$\frac{(t_1)^2}{(t_2)^2} = \frac{Sin(\frac{\theta}{2})}{Sin(\frac{\alpha}{2})}$$

$$\frac{t_1}{t_2} = \frac{\sqrt{Sin(\frac{\theta}{2}}}{\sqrt{Sin(\frac{\alpha}{2}}}$$

That's it. That's the wall I hit. I don't know what to do anymore. Can someone help?

• Why do you want more? You were asked to find the value for the ration$t_1:t_2$ and you got a value for that ration. Commented Mar 8, 2017 at 2:35

Using geometry you can get for the distance AC :
$S=2r\cos\alpha=2r\sin\beta$
where $\alpha$ is the angle MAC and $\beta=90-\alpha$ is the slope of AC. (The polar equation of a circle of radius $a$ with origin at A is $r=2a\sin\theta$.)

The acceleration down AC is $a=g\sin\beta$. The distance is $S=2r\sin\beta$. So the time of descent is $t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{4r}{g}}$. This is independent of $\beta$ (or equivalently $\theta$). Therefore $t_1=t_2$. More generally, the bead will descend in the same time $\sqrt{\frac{4r}{g}}$ along any chord.

Your calculation is close to success.

$\frac{\theta}{2}=\beta$, the slope of AC. The accelerations $a=g\sin\beta$ along AC and AB are not equal, they depend on the slope $\beta$.

• I don't understand what's wrong with my solution, can you clear it up? I made an edit to the final solution but I forgot to include it here accounting for the acceleration, it looked something like this $\frac{t_1}{t_2} = \frac{\sqrt{a_2Sin(\frac{\theta}{2})}}{\sqrt{a_1Sin(\frac{\alpha}{2})}}$ What's wrong with this? Commented Mar 11, 2017 at 3:49
• Nothing wrong, but you have not finished the calculation. How to $a_1,a_2$ relate to $\theta, \alpha$? Commented Mar 11, 2017 at 3:53
• I didn't know. I was so confused I just knew I had to account for acceleration so I put the acceleration placeholders from the kinematic equation in my ratio but I didn't know how to solve for the accelerations. They are just placeholders that I knew were of importance. How do I finish my solution off? edit: I think I now know. I have to account for acceleration due to gravity component acting on the body instead of $a_1$ and $a_2$ and they'll cancel out and give the same ratio. Is this correct? Commented Mar 11, 2017 at 3:54