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Grade 12Mechanics

A car of mass M travles along the arc of a circle of radius R =1 m on a horizontal ground The car starts from rest at time t = 0 and developes a speed V = \sqrt{S} where S = the arc distance traced by the car.
The car starts slipping at time t =2 secs.If the coefficient of static friction between the tyres and the road is \mu, find the value of
\frac{1}{5\mu^{2}}
GUYS PLZ HELP ME OUT

Profile image of Ankit Jaiswal
9 Years agoGrade 12
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To solve this problem, we need to analyze the motion of the car as it travels along the circular path and determine the conditions under which it starts to slip. Let's break this down step by step.

Understanding the Forces at Play

When a car moves in a circular path, it experiences a centripetal force that keeps it on that path. This force is provided by the friction between the tires and the road. The maximum static frictional force can be calculated using the formula:

  • F_friction = μ_s * N

Here, μ_s is the coefficient of static friction, and N is the normal force. On a horizontal surface, the normal force is equal to the weight of the car, which can be expressed as:

  • N = M * g

Where g is the acceleration due to gravity (approximately 9.81 m/s²).

Calculating the Required Centripetal Force

As the car moves along the circular path, it develops a speed V that is related to the distance S it has traveled along the arc. The relationship between the speed and the distance can be expressed as:

  • V = k * S

Where k is a constant that relates speed to distance. The centripetal force required to keep the car moving in a circle is given by:

  • F_c = (M * V²) / R

Substituting the expression for V into the centripetal force equation gives:

  • F_c = (M * (k * S)²) / R

Condition for Slipping

The car will start to slip when the required centripetal force exceeds the maximum static frictional force. Therefore, we set up the inequality:

  • (M * (k * S)²) / R > μ_s * (M * g)

We can simplify this by canceling out M (assuming it's not zero) and rearranging the terms:

  • (k * S)² > μ_s * g * R

Finding the Value of the Coefficient of Static Friction

Now, we know that the car starts slipping at time t = 2 seconds. We can find the distance S traveled in that time using the equation of motion:

  • S = 0.5 * a * t²

Assuming the car accelerates uniformly, we can express the acceleration a in terms of the speed and time. If we know the value of k, we can substitute it back into our inequality to find μ_s.

At the moment of slipping, we can plug in the values:

  • S = 0.5 * a * (2)² = 2a

Substituting this back into our inequality gives:

  • (k * 2a)² > μ_s * g * R

From here, you can solve for μ_s once you have the values for k and a. If you have specific values for these variables, please provide them, and we can calculate the coefficient of static friction together.