Question

A car travels on a flat circular track of 200m at 30m/s and centripetal acceleration is 4.5m/s^2 if the car has a mass of 1000 kg find the frictional force required
If it’s coefficient of static friction is 0.8 what is the maximum speed at which the car can circle the track

Answer 1

Given :

• A car travels on a flat circular track of 200m at 30m/s

• centripetal acceleration = 4.5m/s^2

• mass of car = 1000kg

To find :

• the frictional force required

• If its coefficient of static friction is 0.8 what is the maximum speed at which the car can circle the track

Formula used:

• Force (F) = mass × Acceleration

• Maximum speed (v) = √(μrg)

where :-

• μ = coefficient of static friction
• r = radius
• g = 10 m/s²

Solution :

⟹mass = 1000kg

⟹Acceleration = 4.5m/s²

⟹Friction Force (F) = mass × Acceleration

⟹F = 1000 × 4.5

⟹F = 4500 N

Now we will find maximum speed at which the car can circle the track :-

⟹ v = √(μrg)

Now put :-

• μ = 0.8
• r = 200
• g = 10

⟹$v = \sqrt{0.8 \times 200 \times 10}$

⟹$v = \sqrt{ \dfrac{8}{10} \times 200 \times 10}$

⟹$v = \sqrt{ {8} \times 200 }$

⟹$v = \sqrt{ 1600 }$

⟹$v = \sqrt{ 40 \times 40 }$

⟹$v = 40 \frac{m}{s}$

Answer :

• frictional force = 4500N
• maximum speed = 40 m/s

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Learn more :-

• v = wr

• ar = r w²

• F = mv²/r

• Maximum speed (v) = √(μrg)

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Answer 2

$\bigstar\huge\boxed{\mathtt{\red{Solution:}}}$

As per the given data ,

• The radius of the flat circular track = 200 m
• speed of the car = 30 m/s
• Centripetal acceleration = 4.5 m/s^2
• Mass of the car = 1000 kg
• Coefficient of static friction = 0.8

----------------------------

we know that ,

$\implies\boxed{\mathtt{\green{f = ma }}}$

As the values of mass and acceleration are already given to us we just need to substitute them in the above formula in order to get the answer

$\implies\mathtt{f = 1000 \times 4.5 }$

$\implies\mathtt{f = 450 N }$

The value of the frictional force is 450 N

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The necessary centripetal force for taking turn is provided by the friction between the tyres and the road surface

$\implies\mathtt{F_c= f}$

we know that ,

$\implies\boxed{\mathtt{\blue{F_c=\dfrac{mv^{2} }{r} }}}$

$\implies\boxed{\mathtt{\blue{f = ma =\mu N}}}$

Now substituting the values of Fc and f we get ,

$\implies\mathtt{\dfrac{mv^{2} }{r} = \mu N}$

$\implies\mathtt{\dfrac{mv^{2} }{r} = \mu mg}$

$\implies\mathtt{v^{2} = \mu rg}$

$\implies\mathtt{v^{2} = 0.8\times 200\times 10}$

$\implies\mathtt{v^{2} = 1600}$

$\implies\mathtt{v = \sqrt{ 1600}}$

$\implies\mathtt{\pink{v = 40\dfrac{m}{s}}}$

The maximum speed at which the car can circle the track is 40 m/s