1
$\begingroup$

I ran into this problem in my mechanics homework. enter image description here

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.

enter image description here

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

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

$\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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Mar 8 '17 at 2:35
0
$\begingroup$

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 $\beta$.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Eyad H.
    Mar 11 '17 at 3:49
  • $\begingroup$ Nothing wrong, but you have not finished the calculation. How to $a_1,a_2$ relate to $\theta, \alpha$? $\endgroup$ Mar 11 '17 at 3:53
  • $\begingroup$ 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? $\endgroup$
    – Eyad H.
    Mar 11 '17 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.