You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$\frac{d v}{d r} = 0 \frac{K_{0} [1 - \frac{2 r}{R}]}{2 \sqrt{K_{0} (1 - \frac{r}{R}) r g}} = 0$
$r = \frac{R}{2} V_{m a x} = \sqrt{K_{0} (1 - \frac{\frac{R}{2}}{R}) \frac{R g}{2}} V_{m a x} = \sqrt{\frac{K_{0} R g}{4}}$

Suggest Corrections

Similar questions

Q. 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

O

of the plane as

k = k_{0} (1 - \frac{r}{R})

, where

k_{0}

is a constant. Find the radius of the circle with the centre at the point along which the cyclist can ride with the maximum velocity. What is this velocity?

Q. A cyclist rides along the circumference of a circular horizontal plane of radius R, the frictional coefficient being dependent only on distance `r' from the centre of the plane as

μ = μ_{0} [1 - \frac{r}{R}]

, where

μ_{0}

is a constant. Find the radius of the circle with the centre at the point along which the cyclist can ride with the maximum velocity.

A long cylindrical rod of radius $R_{1}$ is placed co-axially inside a cylindrical tube of radius $R_{2} (> R_{1})$ filled withviscous liquid and pulled along the axis with a constant velocity. In steady state the velocity gradient $(\frac{d v}{d r})$ depends on the distance from the common axis 'r' as proportional to

Q. The kinetic energy

K

of a particle moving along a circle of radius

R

depends on the distance covered as

K = a s^{2},

where

α

is a constant. The force acting on the particle is

Q. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k=a

s^{2}

where a is a constant. The force acting on the particle is

Join BYJU'S Learning Program

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?