Question

24. The centre of gravity of a loaded taxi is 1.5 m above the
ground and the distance between its wheels is 2 m. What is
the maximum speed with which it can go round an
unbanked curve of radius 100 m without being turned
upside down. What minimum value would the coefficient
of friction is needed at this speed ?
[Ans. 25.56 m s -1,0-67]​

Answer 1

Solution :

$\bullet$ Torque will be acting about the point P due to centripetal force and the weight of the car.

$\bullet$ The torque about P due to centripetal force must be balanced by the torque due to weight of the car.

Note: Check this attachment!

Now:

$\boxed{\sf{ \frac{mv ^{2}_{max}}{R} \times 1.5 = mg \times 1}}$

$\implies$ $\sf{v^{2}_{max} = \frac{10 \times 100}{1.5}}$

$\implies$ $\sf{v^{2}_{max} = \frac{1000}{1.5}}$

$\implies$ $\sf{v_{max} = 25.56\:ms^{-1}}$

And:

$\boxed{\sf{\mu \times m \times g = \frac{mv^{2}_{max}}{R}}}$

$\implies$ $\sf{\mu = \frac{(25.56)^{2}}{10 \times 100}}$

$\implies$ $\sf{\mu = 0.67}$

Therefore:

The maximum speed with which it can go round an unbanked curve of radius 100 m without being turned upside down is $\sf{25.56\:ms^{-1}}$ and the minimum value would the coefficient of friction is needed at this speed is $\sf{0.67}$.

_________________

Answered by: Niki Swar, Goa❤️

Answer 2

Answer:

hey mate this is ur answer may it will help u

this diagram will hlp u to understand it

plz follow me