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Question5. (I) What is the maximum speed with which a 1200−kg1200-\mathrm{kg} car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.65 ? Is this result independent of the mass of the car?

Studdy Solution

STEP 1

1. The car is moving in a circular path on a flat road.
2. The coefficient of static friction between the tires and the road is μs=0.65 \mu_s = 0.65 .
3. The radius of the turn is r=90.0 r = 90.0 meters.
4. The mass of the car is m=1200 m = 1200 kg.

STEP 2

1. Identify the forces acting on the car.
2. Use the formula for centripetal force.
3. Relate static friction to centripetal force.
4. Solve for the maximum speed.
5. Discuss the independence of the result from the mass of the car.

STEP 3

Identify the forces acting on the car: - The gravitational force Fg=mg F_g = mg acts downward. - The normal force N N acts upward, equal in magnitude to Fg F_g . - The static frictional force Ff F_f provides the centripetal force needed to keep the car moving in a circle.

STEP 4

Use the formula for centripetal force:
Fc=mv2r F_c = \frac{mv^2}{r}
where v v is the speed of the car, m m is the mass of the car, and r r is the radius of the turn.

STEP 5

Relate static friction to centripetal force: The maximum static frictional force is given by:
Ff=μsN F_f = \mu_s N
Since N=mg N = mg , we have:
Ff=μsmg F_f = \mu_s mg
For maximum speed, the static frictional force equals the centripetal force:
μsmg=mv2r \mu_s mg = \frac{mv^2}{r}

STEP 6

Solve for the maximum speed v v :
μsmg=mv2r \mu_s mg = \frac{mv^2}{r}
Cancel m m from both sides:
μsg=v2r \mu_s g = \frac{v^2}{r}
Solve for v2 v^2 :
v2=μsgr v^2 = \mu_s g r
Take the square root to find v v :
v=μsgr v = \sqrt{\mu_s g r}
Substitute the given values:
v=0.65×9.8 m/s2×90.0 m v = \sqrt{0.65 \times 9.8 \, \text{m/s}^2 \times 90.0 \, \text{m}}
Calculate v v :
v≈573.3 v \approx \sqrt{573.3}
v≈23.94 m/s v \approx 23.94 \, \text{m/s}

STEP 7

Discuss the independence of the result from the mass of the car: The mass m m cancels out in the equation μsmg=mv2r \mu_s mg = \frac{mv^2}{r} , showing that the maximum speed is independent of the mass of the car.
The maximum speed with which the car can round the turn is approximately:
23.94 m/s \boxed{23.94 \, \text{m/s}}

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