A motorcyclist of mass m is to negotiate a cu | vedclass

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

Question

(B) For a motorcyclist to negotiate a curve of radius $r$ with speed $v$ safely,the necessary centripetal force is provided by the static friction bet

Vedclass · Accepted Answer

$\frac{v^2}{gr}$

Vedclass · Answer

(B) For a motorcyclist to negotiate a curve of radius $r$ with speed $v$ safely,the necessary centripetal force is provided by the static friction between the tires and the road.The required centripetal force is $F_c = \frac{mv^2}{r}$.The maximum available static frictional force is $f_{max} = \mu N$,where $N = mg$ is the normal reaction.For safe negotiation,the frictional force must be at least equal to the required centripetal force:$f_{max} \geq F_c$$\mu mg \geq \frac{mv^2}{r}$$\mu \geq \frac{v^2}{rg}$Therefore,the minimum coefficient of friction is $\mu_{min} = \frac{v^2}{rg}$.

$A$ motorcyclist of mass $m$ is to negotiate a curve of radius $r$ with a speed $v$. The minimum value of the coefficient of friction so that this negotiation may take place safely,is

Similar Questions

The value of $\int_{-1}^{1} (\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}) \, dx$ is

Which of the following is the correct formula for the given polymer?

$\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$ is equal to

If $\omega$ is a complex cube root of unity,then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}$ is equal to

The $P-V$ diagram of $2\,g$ of $He$ gas for the process $A \to B$ is shown. What is the heat given to the gas (in $,P_0V_0$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module