A 1900 car starts from rest and drives around a flat 50--diameter circular track. The forward force provided by the car's drive wheels is a constant 1300 .
A) What is the magnitude of the car's acceleration at ?
B) What is the direction of the car's acceleration at ? Give the direction as an angle from the r-axis.
C) If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?
A) To find the magnitude of the car's acceleration at , we can use the equation for acceleration:
acceleration = net force / mass
The net force is the forward force provided by the car's drive wheels, which is a constant 1300 N. However, we need to convert the force to acceleration by dividing it by the car's mass.
To calculate the car's mass, we need to know its weight. We can find the weight by using the equation:
weight = mass * gravity
where gravity is the acceleration due to gravity, approximately 9.8 m/s^2.
Since the car is at rest and experiencing only the force of gravity, the weight is equal to the normal force. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, it is equal to the weight of the car, which we can denote as:
weight = mass * gravity
Solving for mass:
mass = weight / gravity
Now we can substitute this value of mass back into the equation for acceleration:
acceleration = net force / mass
Therefore, the magnitude of the car's acceleration at is:
acceleration = 1300 N / (weight / gravity)
B) To find the direction of the car's acceleration at , we need to consider the centripetal acceleration. Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always points towards the center of the circle.
In this case, the circular track has a radius of 25 m (half of the diameter). Therefore, the centripetal acceleration is given by the equation:
centripetal acceleration = (velocity^2) / radius
Since the car starts from rest, the initial velocity is 0. Therefore, we can ignore the velocity^2 term in the equation, and the centripetal acceleration is simply 0 / 25, which equals 0. (Note: The direction is towards the center of the circle, but we don't need to calculate an angle since it is 0.)
C) To determine when the car begins to slide out of the circle, we need to consider the maximum friction force that the car's tires can provide. This maximum friction force is dependent on the coefficient of static friction between the rubber tires and the concrete track.
The maximum static friction force can be calculated by multiplying the coefficient of static friction (μs) by the normal force. The normal force is equal to the weight of the car, which we already calculated in part A.
Once we have the maximum static friction force, we need to compare it to the centripetal force required to keep the car in circular motion:
centripetal force = mass * centripetal acceleration
If the maximum static friction force is greater than the centripetal force, the car will not slide. If the maximum static friction force is less than the centripetal force, the car will begin to slide out of the circle.
Therefore, we need to set up the following inequality:
maximum static friction force > centripetal force
Now, we can solve for the time when the car begins to slide out of the circle by converting the inequality into an equation:
μs * (mass * gravity) = mass * (velocity^2) / radius
Since we know the values of the radius, maximum static friction force (determined by the coefficient of static friction), and acceleration due to gravity (approximately 9.8 m/s^2), we can substitute them into the equation to solve for the time when the car begins to slide out of the circle.