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Standard XII

Physics

Radial & Tangential Acceleration for Non Uniform Circular Motion

A cyclist rid...

Question

A cyclist rides along the circumference of a circular horizontal plane of radius $R$ , the friction coefficient being dependent only on distance r from the centre $0$ of the plane as $k = k_{o} (1 - r / R)$ , where $k_{o}$ is a constant. What is the maximum safe velocity?

\frac{1}{2} \sqrt{k_{0} R g}

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- \frac{1}{2} \sqrt{k_{0} R g}

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\frac{1}{3} \sqrt{k_{0} R g}

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- \frac{1}{3} \sqrt{k_{0} R g}

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Solution

The correct option is A $\frac{1}{2} \sqrt{k_{0} R g}$
The maximum frictional force must balance the centrifugal force at its maximum velocity.
$\frac{{m v}^{2}}{r} = μ m g v = \sqrt{K_{0} (1 - \frac{r}{R}) r g}$
For velocity to be maximum
$\frac{d v}{d r} = 0 \frac{K_{0} [1 - \frac{2 r}{R}]}{2 \sqrt{K_{0} (1 - \frac{r}{R}) r g}} = 0$
$r = \frac{R}{2} V_{m a x} = \sqrt{K_{0} (1 - \frac{\frac{R}{2}}{R}) \frac{R g}{2}} V_{m a x} = \sqrt{\frac{K_{0} R g}{4}}$

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A cyclist rides along the circumference of a circular horizontal plane of radius R, the friction coefficient being dependent only on distance r from the centre 0 of the plane as k=ko(1−r/R), where ko is a constant. What is the maximum safe velocity?

The correct option is A 12√k0RgThe maximum frictional force must balance the centrifugal force at its maximum velocity.mv2r=μmgv=√K0(1−rR)rgFor velocity to be maximumdvdr=0K0[1−2rR]2√K0(1−rR)rg=0r=R2Vmax=  ⎷K0(1−R2R)Rg2Vmax=√K0Rg4

A cyclist rides along the circumference of a circular horizontal plane of radius $R$ , the friction coefficient being dependent only on distance r from the centre $0$ of the plane as $k = k_{o} (1 - r / R)$ , where $k_{o}$ is a constant. What is the maximum safe velocity?