Question

A particle of mass $m$ moving with constant speed $u$ inside a fixed smooth spherical bowl of radius $a$ describes a horizontal circle at a distance $a/2$ below its centre, then -

• A. the radius of the circular motion is $\frac{a\sqrt{3}}{2}$.
• B. the value of $u$ is $\sqrt{\frac{3ag}{2}}$.
• C. the normal reaction of the spherical surface on the particle is $\frac{mg}{2}$.
• D. the magnitude of the resultant force acting on the particle is zero, in an inertial force.

Answer 1

Answer: (A), (B)

the value of $u$ is $\sqrt{\frac{3ag}{2}}$.

Let us represent the given situation diagrammatically.

Now applying Pythagoras theorem in $△OAB$, we have,

$a^2={\left(\frac{a}{2}\right)}^2+d^2,$ where $d$ is the radius of the circular motion.

$\Rightarrow d^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$

$\Rightarrow d=\frac{\sqrt{3}a}{2}$

Further, balancing force in vertical direction, we get,
$N\sin \theta =mg$

$\Rightarrow N\times \frac{a/2}{a}=mg$

$\Rightarrow N=2mg$

Also, from horizontal direction, we get,
$N\cos \theta =\frac{m{u}^2}{d}$

$\Rightarrow N\times \frac{\sqrt{3}a/2}{a}=\frac{m{u}^2}{\left(\sqrt{3}a/2\right)}$

$\Rightarrow 2mg\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}a}{2}=m{u}^2$

$\Rightarrow u=\sqrt{\frac{3ag}{2}}$

The magnitude of force is not zero in inertial frame, since the particle is performing circular motion $i.e.$ accelerated motion.