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. $2 a \frac{ s ^{2}}{ R }$
• B. $2 as \left(1+\frac{ s ^{2}}{ R }\right)^{1 / 2}$
• C. $2 as$
• D. $2 a \frac{ R ^{2}}{ s }$

Answer 1

Answer: (B)

$2 as \left(1+\frac{ s ^{2}}{ R }\right)^{1 / 2}$

Answer 2

In non-uniform circular motion two forces will work on a particle $F _{ c }$ and $F _{ t }$
So, the net force $F _{ Net }=\sqrt{ F _{ c }^{2}+ F _{ t }^{2}}$ ... (i)
Centripetal force $F _{ c }=\frac{ mv ^{2}}{ R }=\frac{2 as ^{2}}{ R }$ ... (ii)
$\left[\right.$ Given $\left.\frac{1}{2} mv ^{2}= as ^{2}\right]$
Again from $\frac{1}{2} mv ^{2}= as ^{2}$
$\Rightarrow v ^{2}=\frac{2 as ^{2}}{ m }$
$\Rightarrow v = s \sqrt{\frac{2 a }{ m }}$
Tangential acceleration
$a _{ t }=\frac{ dv }{ dt }=\frac{ dv }{ ds } \cdot \frac{ ds }{ dt }$
$a_{t}=v \sqrt{\frac{2 a}{m}}=\frac{2 a s}{m}$
and $F _{ t }= ma _{ t }=2$ as...(iii)
Now on substituting value of $F_{c}$ and $F_{t}$ in E (i) we get
$\therefore F _{ Net }=\sqrt{\left(\frac{2 as ^{2}}{ R }\right)^{2}+\left(2 as ^{2}\right)^{2}}=2 as$
$=2 as \left(1+\frac{ s ^{2}}{ R }\right)^{1 / 2}$