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PhysicsW.E.P.Vertical circular motionEasy2 minPYQ_2020

PhysicsEasysingle choice

What will be the force acting on a particle moving along a circle of radius $R$ , if its kinetic energy $k$ depends on the distance covered as $k = a s^{2}$ ?

Options:

Answer:

Solution:

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

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

$∴ a_{R} = \frac{v^{2}}{R} = \frac{2 a s^{2}}{m R}$ .... (i)

$a_{t} = \frac{d v}{d t} = \frac{d v}{d s} . \frac{d s}{d t} = v \frac{d v}{d s}$ .... (ii)

From equation (i)

$a_{t} = [s \sqrt{\frac{2 a}{m}}] [\sqrt{\frac{2 a}{m}}] = \frac{2 a s}{m}$ ..... (iii)

So, net accelerations $= \sqrt{a_{R}^{2} + a_{t}^{2}}$

${= \sqrt{{[\frac{2 a s^{2}}{m R}]}^{2} + {[\frac{2 a s}{m}]}^{2}}}^{}$

Hence, $a_{0} = \frac{2 a s}{m} \sqrt{1 + {[\frac{s}{R}]}^{2}}$

$∴ F = m a_{0} = 2 a s \sqrt{1 + {(\frac{s}{R})}^{2}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion

⏱ 2mℹ️ Source: PYQ_2020

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