1. ## Mechanics MC help

http://prntscr.com/ftv4dg

Answer is C, any help appreciated.

2. ## Re: Mechanics MC help

Originally Posted by pikachu975
http://prntscr.com/ftv4dg

Answer is C, any help appreciated.
Quick answer: C is the only option with the right units (or dimensions), so it's the answer.

(As we know, the dimensions of angular velocity are inverse time (units of which is s-1); units of option C is ((m s-2)/m)1/2 = s-1. It is quick to check that none of the other options have the right units for angular velocity (and so cannot be correct). The first two (A and B) are dimensionless, as they just involve the ratio of two lengths. The last option is the inverse of C, so the units are s, which is wrong.)

3. ## Re: Mechanics MC help

Originally Posted by InteGrand
Quick answer: C is the only option with the right units (or dimensions), so it's the answer.

(As we know, the dimensions of angular velocity are inverse time (units of which is s-1); units of option C is ((m s-2)/m)1/2 = s-1. It is quick to check that none of the other options have the right units for angular velocity (and so cannot be correct). The first two (A and B) are dimensionless, as they just involve the ratio of two lengths. The last option is the inverse of C, so the units are s, which is wrong.)
Thanks heaps! Didn't realise radians per second was the same as s^-1. Also is there a method to doing it if this was a short answer?

4. ## Re: Mechanics MC help

Originally Posted by pikachu975
Thanks heaps! Didn't realise radians per second was the same as s^-1. Also is there a method to doing it if this was a short answer?
$\noindent The reason that \omega still has dimensions of inverse time is that angles are dimensionless (being the ratio of two lengths). And you could do this question using usual methods (resolving forces to get simultaneous equations, remembering that the centripetal force is mr\omega^{2}, and \tan \alpha = \frac{r}{h}).$

$\noindent In other words, resolving forces gives us the system of equations$

$\begin{cases}T\cos \alpha = mg \quad (1) \\ T\sin \alpha = m r \omega^{2}.\quad (2)\end{cases}$

$\noindent Dividing (2) by (1) shows that$

$\tan \alpha = \frac{r\omega^{2}}{g}\Rightarrow \omega = \sqrt{\frac{g}{r}\tan\alpha}.$

$\noindent Using \tan \alpha = \frac{r}{h} (which follows from right-angle trigonometry), we thus have \omega = \sqrt{\frac{g}{h}}.$

5. ## Re: Mechanics MC help

Originally Posted by InteGrand
$\noindent The reason that \omega still has dimensions of inverse time is that angles are dimensionless (being the ratio of two lengths). And you could do this question using usual methods (resolving forces to get simultaneous equations, remembering that the centripetal force is mr\omega^{2}, and \tan \alpha = \frac{r}{h}).$

$\noindent In other words, resolving forces gives us the system of equations$

$\begin{cases}T\cos \alpha = mg \quad (1) \\ T\sin \alpha = m r \omega^{2}.\quad (2)\end{cases}$

$\noindent Dividing (2) by (1) shows that$

$\tan \alpha = \frac{r\omega^{2}}{g}\Rightarrow \omega = \sqrt{\frac{g}{r}\tan\alpha}.$

$\noindent Using \tan \alpha = \frac{r}{h} (which follows from right-angle trigonometry), we thus have \omega = \sqrt{\frac{g}{h}}.$
Thanks, didn't know we were allowed to use horizontal forces because the question was like "taking into account only tension and gravity"

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