Ball on a string

Classical Mechanics Level 3

A $5\text{ kg}$ object is tied to the end of a $0.2\text{ m}$ string and swung in a horizontal circle while making an angle of $\frac{\pi}{6}$ below the horizontal. If $g=10\frac{\text{m}}{\text{s}^2},$ how fast is the object traveling in $\frac{\text{m}}{\text{s}}?$

2

\sqrt{3}

1

2\sqrt{3}

1 solution

July Thomas
May 11, 2016

$\Sigma F_y = F_T \sin(\frac{\pi}{6}) - mg = 0$ $F_T = \frac{mg}{\sin(\frac{\pi}{6})}$ $F_T = \frac{(5)(10)}{\frac12}$ $F_T = 100\text{ N}$

$\Sigma F_x = F_T\cos(\frac{\pi}{6}) = m\frac{v^2}{r}$ $v = \sqrt{\frac{F_T r\cos(\frac{\pi}{6})}{m}} = \sqrt{\frac{(100) \big(0.2\cos (\frac{\pi}{6}) \big) (\frac{\sqrt{3}}{2})}{5}} = \sqrt{3} \frac{\text{m}}{\text{s}}$

Hi July,

I really like your problem. I was just about to post the exact same one, then I saw yours!

However, when I worked your problem, I did not arrive at an answer which is one or your selections. I am looking at your solution, and we agree up to the last two lines. Let me write out what I have and then we can discuss, okay?

$\sum F_x = F_T \cos(\pi/6) = m \frac{v^2}{r \cos(\pi/6)}$

$v = \sqrt{\frac{F_T r \cos^2 (\pi/6)}{m}} = \sqrt{3} \text{m/s}$

I put a factor of $\cos(\pi/6)$ in my radius because this is the true radius of the object's circular trajectory. I might be off on this, but what do you think?

Aaron Miller Staff - 5 years ago

Hey Aaron, thank you for pointing that out! My radius seemingly disappeared, which is usually not permitted in algebra. I have amended my solution.

July Thomas - 4 years, 12 months ago

Ball on a string

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