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Work and Energy — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork and Energy5 Marks
Question
A particle of mass m is kept on a fixed, smooth sphere of radius R at a position, where the radius through the particle makes an angle of 30° with the vertical. The particle is released from this position.
  1. What is the force exerted by the sphere on the particle just after the release?
  2. Find the distance travelled by the particle before it leaves contact with the sphere.
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Answer

  1.  


When the particle is released from rest (fig), the centrifugal force is zero. N force is zero $=\text{mg}\cos\theta$

$=\text{mg}\cos30^\circ=\frac{\sqrt3\text{mg}}{2}$
  1.  


When the particle leaves contact with the surface (fig), N = 0.

So, $\frac{\text{mv}^2}{\text{R}}\text{mg}\cos\theta$

$\Rightarrow\text{v}^2=\text{Rg}\cos\theta\ \dots(1)$

Again, $\frac{1}{2}\text{mv}^2=\text{mgR}(\cos30^\circ-\cos\theta)$

$\Rightarrow\text{v}^2=2\text{Rg}\Big(\frac{\sqrt3}{2}-\cos\theta\Big)\ \dots(2)$

From equn. (1) and equn. (2),

$\text{Rg}\cos\theta=\sqrt3\text{Rg}-2\text{Rg}\cos\theta$

$\Rightarrow3\cos\theta=\sqrt3$

$\Rightarrow\cos\theta=\frac{1}{\sqrt3}$

$\Rightarrow\theta=\cos^{-1}\frac{1}{\sqrt3}$

So, the distance travelled by the particle before leaving contact,

$\ell=\text{R}\Big(\theta-\frac{\pi}{6}\Big)$ $\Big[\text{because}30^\circ=\frac{\pi}{6}\Big]$

putting the value of $\theta,$ we get $\ell=0.43\text{R}$

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