The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. It is given as KE = as 2, where a is a constant. The force acting on the particle is

Question

The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. It is given as KE = as 2, where a is a constant. The force acting on the particle is (A) 2 a ( s 2/ R ) (B) 2 as (1+( s 2/ R ))1 / 2 (C) 2 as  (D) 2 a ( R 2/ s )

Tardigrade Admin · Accepted Answer

In non-uniform circular motion two forces will work on a particle

F_{c}

and

F_{t}

So, the net force

F_{N e t} = F_{c}^{2} + F_{t}^{2}

... (i)
Centripetal force

F_{c} = \frac{m v ^{2}}{R} = \frac{2 a s ^{2}}{R}

... (ii)

[

Given

\frac{1}{2} m v^{2} = a s^{2}]

Again from

\frac{1}{2} m v^{2} = a s^{2}

\Rightarrow v^{2} = \frac{2 a s ^{2}}{m}

\Rightarrow v = s \frac{2 a}{m}

Tangential acceleration

a_{t} = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t}

a_{t} = v \frac{2 a}{m} = \frac{2 a s}{m}

and

F_{t} = m a_{t} = 2

as...(iii)
Now on substituting value of

F_{c}

and

F_{t}

in Eq. (i) we get

∴ F_{N e t} = (\frac{2 a s ^{2}}{R})^{2} + (2 a s^{2})^{2} = 2 a s

= 2 a s (1 + \frac{s ^{2}}{R})^{1/2}

Q. The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E = a s^{2}$ , where a is a constant. The force acting on the particle is

$2 a \frac{s ^{2}}{R}$

$2 a s (1 + \frac{s ^{2}}{R})^{1/2}$

$2 a s$

$2 a \frac{R ^{2}}{s}$

Q. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. It is given as KE=as2, where a is a constant. The force acting on the particle is

2aRs2​

2as(1+Rs2​)1/2

2as

2asR2​

Q. The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E = a s^{2}$ , where a is a constant. The force acting on the particle is

$2 a \frac{s ^{2}}{R}$

$2 a s (1 + \frac{s ^{2}}{R})^{1/2}$

$2 a s$

$2 a \frac{R ^{2}}{s}$