Question

A toy car is moving on a circular path of radius 10 metre on an inclined plane of inclination 30 degree the coefficient of friction between the tyres and inclined plane is 2/√3. find maximum constant speed of car such that car does not slip. assume that the width of car is negligible wrt radius of circular

Answer 1

Answer:

Vmax=(rg(f+tanx)/1-tanx)^1/2

Explanation:

r=10  x=30

f=2/$\sqrt{3$ g=10

Vmax=22.8

Answer 2

Answer:

The maximum constant speed of car such that car does not slip is 5.8m/s.

Explanation:

Constructing a free body diagram for this car.

In this F.B.D,

coefficient of friction: $\mu = \frac{2}{\sqrt{3} }$,

radius: $r=10m$.

$\frac{mv^2}{r}$ is the centrifugal force,    [where $v$ is the required constant velocity]

$N$ is the normal reaction.

Now, making components of $\mu N$ and $N$.

Now we can equate horizontal forces, as there is no horizontal motion.

$\implies \frac{mv^2}{r}+\mu N cos(30^o)=Nsin(30^o)$

$\implies \frac{mv^2}{10}+\frac{2}{\sqrt{3} } N \frac{\sqrt{3} }{2 } =N\frac{1}{2}$

$\implies \frac{mv^2}{10}=\frac{N}{2}$

$\implies v=\sqrt{\frac{5N}{m}}$

we can equate vertical forces, as there is no vertical motion.

$\implies \mu Nsin(\theta)+Ncos(\theta)=mg$

$\implies \frac{2}{\sqrt{3} } *N*\frac{1}{2} +N*\frac{\sqrt{3} }{2} =mg$

$\implies \frac{N}{\sqrt{3} }+\frac{\sqrt{3} }{2}N =mg$

$\implies N =\frac{2\sqrt{3} mg}{5}$

Putting it in the value of $v$, we get,

$\implies v=\sqrt{\frac{5(\frac{2\sqrt{3} mg}{5})}{m}}$

$\implies v=\sqrt{2\sqrt{3} g }=\sqrt{2\sqrt{3}} *9.8$

$\implies v=5.8m/s$

The maximum constant speed of car such that car does not slip is 5.8m/s.

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