A car 1500 kg is turning a flat curve while moving at constant speed. The radius of the curve is 35.0 m. If the coefficient of static friction between tires and dry pavement is 0.523. Determine the maximum speed of car to negotiate the turn successfully?

Answer 1

The centripetal force required for circular motion of the car is provided by the force of friction between the tires and pavement. Due to Newton's Third law of motion, the car experiences centrifugal force which tends to topple it off the road in the outward direction.

Force of friction along the radius #R# is given as
# f_r= mu_s|vecN| # where #vecN# is normal reaction and #=mvecg#, where #m# is mass of car and #g# is gravity #=9.81\ ms^-2#
When the car having velocity #v_max# is on the verge of skidding force of friction must be equal to centripetal force which is given as
#F_c=(mv_max^2)/R#

Equating the two and inserting given values we get

#(mv_max^2)/R=mu_smg # #=>v_max^2=mu_sg R# #=>v_max =sqrt((0.523)(9.81)(35.0))# #=>v_max=13.4\ ms^-1#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To determine the maximum speed of the car to negotiate the turn successfully, we can use the formula for centripetal force:

[ F_c = \frac{mv^2}{r} ]

Where:

  • ( F_c ) is the centripetal force,
  • ( m ) is the mass of the car (1500 kg),
  • ( v ) is the velocity of the car (the maximum speed we want to find),
  • ( r ) is the radius of the curve (35.0 m).

The maximum centripetal force that can be exerted by the static friction is given by:

[ F_{\text{max}} = \mu_s \cdot N ]

Where:

  • ( \mu_s ) is the coefficient of static friction (0.523),
  • ( N ) is the normal force exerted on the car due to gravity.

We can find the normal force ( N ) using:

[ N = mg ]

Where:

  • ( g ) is the acceleration due to gravity (9.8 m/s²).

Substituting the expressions for ( F_c ) and ( F_{\text{max}} ) and solving for ( v ), we get:

[ \frac{mv^2}{r} = \mu_s \cdot N ]

[ \frac{mv^2}{r} = \mu_s \cdot mg ]

[ v^2 = \mu_s \cdot g \cdot r ]

[ v = \sqrt{\mu_s \cdot g \cdot r} ]

Plugging in the given values:

[ v = \sqrt{0.523 \cdot 9.8 \cdot 35} ]

[ v = \sqrt{181.547} ]

[ v \approx 13.47 , \text{m/s} ]

So, the maximum speed of the car to negotiate the turn successfully is approximately 13.47 m/s.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7