Question

To test the performance of its tires, a car

travels along a perfectly flat (no banking) cir-

cular track of radius 516 m. The car increases

its speed at uniform rate of

at ≡d |v|÷dt = 3.89 m/s^2

until the tires start to skid.

If the tires start to skid when the car reaches

a speed of 32.8 m/s, what is the coefficient of

static friction between the tires and the road? The acceleration of gravity is 9.8 m/s^2​

Answer 1

The coefficient of static friction (μ_s) can be found using the formula μ_s = v^2/(gr), where v is the speed of the car at the point of skidding, g is the acceleration due to gravity, and r is the radius of the track. Substituting in the provided values, the minimum coefficient of static friction needed to prevent the car from skidding is approximately 0.207.

Step-by-step explanation:

To find the coefficient of static friction between the tires and the road, we can use the formula for the force due to friction, which is the product of the coefficient of static friction (μ_s) and the normal force (N). In this scenario, the normal force is equal to the gravitational force because the track is flat, so N = mg, where m is the mass of the car and g is the acceleration of gravity. The centripetal force needed to keep the car moving in a circle is F_c = mv^2/r, where v is the velocity of the car and r is the radius of the circular track.

Since the car starts to skid at a certain speed, this means that the centripetal force has exceeded the maximum force of static friction, hence F_c = μ_s * N. By plugging the values we have into this equation, we can solve for the coefficient of static friction:

F_c = mv^2/rN = mgF_c = μ_s * N

From this, we get μ_s = (mv^2)/(mgr), meaning that the mass of the car cancels out, leaving us with:

μ_s = v^2/(gr)

Using the given values, v = 32.8 m/s, g = 9.8 m/s^2, and r = 516 m, we obtain:

μ_s = (32.8 m/s)^2/((9.8 m/s^2)(516 m))

After calculating, we find the minimum coefficient of static friction required to prevent skidding is approximately:

μ_s = 0.207

Answer 2

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration
`a_r`= 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

`a_{R = (v^2)/(r)`

`a_{R = ((32.8)^2)/(516)`

`a_{R = ((1075.84))/(516)`

`a_{R = 2.085 m/s^2`

Now the total acceleration is

`\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2`

=>
`= √( (a_r)^2+(a_R)^2)`

=>
`√( (3.89 )^2+( 2.085)^2)`

=>
`√( (15.1321)+(4.347)^2)`

=>
`19.4791 m/s^2`

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

`\mu =(f)/(W)`

From (1) and (2)

`\mu =(ma)/(mg)`

`\mu =(a)/(g)`

Substituting the values, we get

`\mu =(19.4791)/(9.8)`

`\mu =1.987`