Question

A pendulum has a ball of mass m attached to the string and is suspended from the roof of a trolley. If the trolley rolls upwards with acceleration a, then what is the angle made by the string with the inclined plane? (Given that a = $5m/s^2,g=10m/s^2$ and the angle of inclination of the plane is ${30}^{\circ }$

• A. $co{s}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
  
• B. $co{t}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
  
• C. $si{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
  
• D. $ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is B $co{t}^{-1}\left(\frac{2}{\sqrt{3}}\right)$



In equilibrium,
$Tcos\theta =ma+mgsin\alpha$
And $Tsin\theta =mgcos\alpha$
This gives,
$cot\theta =\frac{a+gsin\alpha }{gcos\alpha }=\frac{5+10\times \left(\frac{1}{2}\right)}{10\times \frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}\therefore =co{t}^{-1}\left(\frac{2}{\sqrt{3}}\right)$