The kinetic energy k of a particle moving alo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given kinetic energy $k = \frac{1}{2}mv^2 = as^2$. From this,the velocity is $v = s\sqrt{\frac{2a}{m}}$. The centripetal acceleration is $a_R = \f

Vedclass · Accepted Answer

$2as\sqrt{1 + \frac{s^2}{R^2}}$

Vedclass · Answer

(B) Given kinetic energy $k = \frac{1}{2}mv^2 = as^2$. From this,the velocity is $v = s\sqrt{\frac{2a}{m}}$. The centripetal acceleration is $a_R = \frac{v^2}{R} = \frac{2as^2}{mR}$. The tangential acceleration is $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$. Substituting $v$,we get $a_t = \left(s\sqrt{\frac{2a}{m}}\right) \frac{d}{ds}\left(s\sqrt{\frac{2a}{m}}\right) = s\left(\frac{2a}{m}\right) = \frac{2as}{m}$. The net acceleration is $a_{net} = \sqrt{a_R^2 + a_t^2} = \sqrt{\left(\frac{2as^2}{mR}\right)^2 + \left(\frac{2as}{m}\right)^2}$. Simplifying,$a_{net} = \frac{2as}{m} \sqrt{\frac{s^2}{R^2} + 1}$. The net force is $F = m \cdot a_{net} = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

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

Similar Questions

In non-uniform circular motion,the ratio of tangential to radial acceleration is ($r=$ radius,$\alpha=$ angular acceleration,$V=$ linear velocity).

$A$ particle of mass $5 \ kg$ is moving on a circle of radius $1 \ m$ with an angular velocity of $2 \ rad \ s^{-1}$. The centripetal force is (in $N$)

The magnitude of the centripetal force acting on a body of mass $m$ executing uniform circular motion in a circle of radius $r$ with speed $v$ is:

The distance of a particle moving on a circle of radius $12 \, m$ measured from a fixed point on the circle and measured along the circle is given by $s = 2t^3$ (in meters). The ratio of its tangential to centripetal acceleration at $t = 2 \, s$ is .........

$A$ car is travelling with linear velocity '$V$' on a circular road of radius '$r$'. If its velocity is increasing at a rate of '$a$' $ms^{-2}$, then the resultant acceleration will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module