Find the least possible value of U^2?

Question

QS:A particle P of mass m is connected to a ﬁxed point O by a light inextensible string OP of length r, and is moving in a vertical circle, centre O. At its lowest point, P has speed U. When the string makes an angle α with the downward vertical it encounters a small ﬁxed peg Q, where OQ=1 2 r. The string proceeds to wrap itself around the peg, so that P begins to move in a vertical circle with centre Q (see diagram). Given that the particle describes a complete circle about Q, show that U^2≥gr(7/2−cosα) .

My Thoughts: I used the energy conservation to find the inequality for U^2 taking into consideration that the final velocity is zero at the highest point. How to find the vertical distance between the lowest and highest point?? I found the distance between lowest point and pint string just before touching the peg using trigonometry easily. PLEASE HELP!!

What is the inequality you are trying to prove? – David Quinn May 07 '18 at 20:55 — David Quinn, May 07 '18 at 20:55
U^2≥gr7/2−cosα . – Powerful blaster May 08 '18 at 08:47 — Powerful blaster, May 08 '18 at 08:47

Mark Fischler · Answer 1 · 2018-05-13T02:01:01.547

The key insight for this problem is that at the top of the circle about $Q$, the mass cannot be travelling at zero circular velocity; if it were, then instead of completing the circle, it would just fall straight toward $Q$. In fact, it must be travelling fast (angular velocity $\omega$ enough that the "centrifugal force" at least balances the gravitational force): $$ m\omega^2 (\frac12 r) \geq mg \\ \omega \geq \sqrt{\frac{2g}{r}} \\ v = \left(\frac12 r \right) \omega \geq \frac12 \sqrt{2gr} $$ and the kinetic energy at the top of the loop is given by $$ E_k = \frac12 mv^2 \geq \frac14 mgr $$ Now the height at the top of the loop is the height of the peg at $Q$ plus the $\frac{r}{2}$ length of the wrapping string: $$ h = \left( r- \frac12 r \cos \alpha \right) + \frac12 r = \frac32 r - \frac12 r \cos \alpha $$ So the total energy at the top, which can't be more that the initial kinetic energy at the bottom due to the bottom-speed $u$, is
$$ E_{\mbox{tot}} \geq \frac14 mgr + mg\left( \frac32 r - \frac12 r \cos \alpha \right)$$ $$ \frac12 m u^2 \geq \frac12 mg \left( \frac12 r + \color{red}{\frac32 r - \frac12 r \cos \alpha} \right) $$ $$ u^2 \geq gr\left( 2-\frac12\cos\alpha\right) $$ EDIT The expression in red is incorrect; since we have factored out $\frac12 mg$, the equations from there should read $$ \frac12 m u^2 \geq \frac12 mg \left( \frac12 r + \color{red}{3 r - r \cos \alpha} \right) $$ $$ u^2 \geq gr\left( \color{red}{\frac72 -\cos\alpha}\right) $$

Note that this makes the answer given by @Cesar Eo the correct answer.

...Actually that's not correct. The necessary condition for a complete circle is that there is at least some tension in the string at the top of the circle about $Q$. — David Quinn, May 07 '18 at 20:10

score 0 · Accepted Answer · answered May 08 '18 at 13:47

Along the circular path in the small circle we should have

$$ T = m g\cos\beta + m \frac{2 v_{\beta}^2}{r} > 0 \Rightarrow v_{\beta}^2 > -\frac{r g}{2}\cos\beta\Rightarrow v_{\pi}^2 \gt \frac{rg}{2} $$

when $\beta = \pi$ (vertical). At this point we have

$$ h_{\pi} = \frac{r}{2}(3-\cos\alpha) $$

and

$$ \frac{1}{2}mU^2 = \frac{1}{2}m v_{\pi}^2+m h_{\pi}g $$

or

$$ v_{\pi}^2 = U^2-2gh_{\pi} \gt \frac{rg}{2} $$

hence

$$ U^2 \gt r g \left(\frac{7}{2}-\cos\alpha\right) $$

Find the least possible value of U^2?

2 Answers2