Tag Archives: Circular Motion

Water in Vertical Circle

Consider a small bucket full of water tied to a string whirled around in vertical circle of radius r without water falling down. At the topmost position when the speed of the inverted bucket is v,

(A) $\frac{{{v^2}}}{r} = g$ necessarily
(B) $\frac{{{v^2}}}{r} < g$ necessarily
(C) $\frac{{{v^2}}}{r} > g$ necessarily
(D) None of the options given

Solution

Consider water as the system. Let, m be the mass of water. The weight mg acts downwards and the normal reaction N also acts downwards.

$mg + N = \frac{{m{v^2}}}{r}$

$ \Rightarrow \frac{{m{v^2}}}{r} \ge mg$

$ \Rightarrow \frac{{{v^2}}}{r} \ge g$

Hence, Option (D).

Centripetal & Centrifugal Forces

In circular motion, why don’t centripetal and centrifugal forces cancel each other? Well, they don’t cancel each other because they are not present simultaneously for the same observer. Only one is present at a time for one observer. If you take ground observer – inertial observer to be more precise – then only centripetal force is applicable and of course it is applicable towards the centre. There is no centrifugal force present here. Since both are not present simultaneously, they don’t cancel each other. In case of non-inertial or accelerated observer, only centrifugal force is present – it acts away from the centre. There is no centripetal force applicable. Since both are not present simultaneously, they can’t cancel each other.

JEE Main 2018 Circular Motion Potential Problem

A particle is moving in a circular path of radius a under the action of an attractive potential $U=-\frac {k}{2r^2}$. Its total energy is:

(1) $-\frac {k}{4a^2}$
(2) $\frac {k}{2a^2}$
(3) Zero
(4) $-\frac {3}{2} \frac {k}{a^2}$

[Based on JEE Main 2018]

Solution

Let the mass of the particle be m,

$P.E.=mU=-\frac {1}{2} \frac {mk}{r^2}$

$T.M.E.=\frac {1}{2} mv^2 +\left(-\frac {1}{2} \frac {mk}{r^2}\right)$

Where v is the speed of the particle.

$T.M.E.=\frac {1}{2} m \left(v^2 – \frac {k}{r^2}\right)$

$F=- \frac {d}{dr} \left(-\frac {1}{2} \frac {mk}{r^2}\right)=-\frac {mk}{r^3}$

Since F provides centripetal force,

$\left| { – \frac{{mk}}{{{r^3}}}} \right| = \frac{{m{v^2}}}{r},\therefore \frac{k}{{{r^2}}} = {v^2}$

Hence, T.M.E. = 0

Hence, Option (3).


A particle is moving in a circular path of radius a under the action of attractive potential U = -k/2r^2. Its total energy is:


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