Question

A particle describes a horizontal circle on the smooth surface of an inverted cone. The plane of that circle is at a height of 9.8 cm above the vertex. Then the speed of the particle is:
A) $0.49m{s^{ - 1}}$
B) $0.98m{s^{ - 1}}$
C) $1.96m{s^{ - 1}}$
D) $3.92m{s^{ - 1}}$

Answer 1

In this question we have to find the value of speed of a particle which is describing a circle at a height of 9.8 cm. to find the value of speed, we will use Newton’s law of motion.

Complete step by step solution:
Given,
Height, $h = 9.8cm$,
If N is the normal to the circle then the $N\cos \theta$and $N\sin \theta$ will be two components of normal N.
$N\sin \theta$ will be equivalent to mg and $N\cos \theta$ will be equivalent to $\dfrac{{m{v^2}}}{r}$.
$\Rightarrow N\sin \theta$= mg …. (1)
$\Rightarrow N\cos \theta$= $\dfrac{{m{v^2}}}{r}$…… (2)
Where, m is mass of the particle
v is velocity and
r is the radius of motion.
Dividing equations (1) and (2)
$\Rightarrow \tan \theta = \dfrac{{mg}}{{m{v^2}/r}}$
The radius of motion $r = h\tan \theta$
Putting the values of$\tan \theta$,
$\Rightarrow r = h\tan \theta$
$\Rightarrow r = h\dfrac{{mg}}{{m{v^2}/r}}$
$\Rightarrow r = \dfrac{{rhg}}{{{v^2}}}$
From above equation (3), we will find the value of ${v^2}$
$\Rightarrow {v^2} = gh$
Putting values of g and h
$\Rightarrow {v^2} = 9.8 \times 9.8 \times {10^{ - 2}}$
$\Rightarrow {v^2} = 96 \times {10^{ - 2}}$
$\Rightarrow {v^2} = 0.96$
$\Rightarrow v = 0.98m{s^{ - 1}}$

Hence, from above calculation we have seen that the speed of the particle is $v = 0.98m{s^{ - 1}}$.

Note: In this question we had to find the speed of a particle moving in a circular motion at a height. In such conditions it is necessary to analyze the situation.as in this case first we found the components of normal to the motion of the particle. Then we used the equation of radius of motion. $\dfrac{{m{v^2}}}{r}$ is the centripetal force acting on the particle. We have equated this force with the parallel component $N\cos \theta$ of normal and the weight of the particle has been equated with the perpendicular component.