4. A 1500 kg car starts from rest and drives around a flat 50 m diameter circular track. The
forward force provided by the car’s wheels is a constant 1000 N.
a. What are the magnitudes and direction of the car’s acceleration at t = 10 s?
b. If the car has rubber tires and the track is concrete (µs = 1.00), at what time does
the car begin to slide?
2850
To determine the car's acceleration at t = 10 s, we need to use the equations of circular motion.
a. Magnitude of the car's acceleration:
The centripetal acceleration (ac) can be calculated using the equation:
ac = (v^2) / r
Given that the car is moving around a circular track with a diameter of 50 m (radius = 25 m), we can calculate the car's velocity (v) at t = 10 s using the formula:
v = at
where a is the car's constant forward force provided by the wheels (a = 1000 N) and t = 10 s.
Substituting the values into the equations:
v = a * t
= 1000 N * 10 s
= 10000 N s
Next, we can calculate the magnitude of centripetal acceleration:
ac = (v^2) / r
= (10000 N s)^2 / 25 m
= 40000000 N^2 s^2 / 25 m
= 1600000 N^2 s^2 / m
Therefore, the magnitude of the car's acceleration at t = 10 s is 1600000 N^2 s^2 / m.
Direction of the car's acceleration:
The acceleration is directed towards the center of the circular track, which is inward.
b. To determine the time at which the car begins to slide, we need to compare the maximum static frictional force to the force required for the car to stay in circular motion. If the maximum static frictional force is exceeded, the car will start sliding.
The maximum static frictional force can be calculated using the equation:
fs(max) = µs * N
where µs is the static friction coefficient (given as 1.00), and N is the normal force.
The normal force (N) can be calculated by summing the gravitational force and the vertical component of the centripetal force:
N = mg + (mv^2) / r
where m is the mass of the car (given as 1500 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), v is the car's velocity, and r is the radius of the circular track (25 m).
Substituting the values into the equations:
N = (1500 kg * 9.8 m/s^2) + (1500 kg * 10000 N s^2 / 25 m) / 25 m
= 14700 N + 600 N
= 15300 N
Now, we can calculate the maximum static frictional force:
fs(max) = µs * N
= 1.00 * 15300 N
= 15300 N
Since the forward force provided by the car's wheels (1000 N) is less than the maximum static frictional force (15300 N), the car will not start sliding.
Therefore, the car will never start sliding on the concrete track.
To find the magnitudes and direction of the car's acceleration at t = 10 s, we need to consider the forces acting on the car.
a. At t = 10 s, the car has been moving for some time, so it won't be at rest anymore. The force provided by the car's wheels is responsible for the car's acceleration. Since the car's speed isn't provided, we need to calculate it first.
To find the speed of the car at t = 10 s, we can use the formula:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration (unknown)
t = time (10 s)
Rearranging the formula, we get:
a = (v - u) / t
Since the car is moving around a circular track, its acceleration is directed towards the center of the circle. This centripetal acceleration can be calculated using the formula:
a = v^2 / r
Where:
a = centripetal acceleration
v = velocity
r = radius of the circle (diameter/2)
Since the car moves in a circle of 50 m diameter, the radius would be 50/2 = 25 m.
Now we can set up an equation using the two equations for acceleration:
(v - u) / t = v^2 / r
Substituting the known values:
(1000 N) / (1500 kg) = v^2 / 25 m
Simplifying, we get:
v^2 = (1000 N) / (1500 kg) * 25 m
v^2 = 33.33 m^2/s^2
Taking the square root of both sides, we get:
v = √(33.33 m^2/s^2)
v ≈ 5.77 m/s
Now we can calculate the acceleration using the centripetal acceleration formula:
a = v^2 / r
a = (5.77 m/s)^2 / 25 m
a ≈ 1.326 m/s^2
So, at t = 10 s, the car's acceleration is approximately 1.326 m/s^2, directed towards the center of the circular track.
b. To determine when the car begins to slide, we need to consider the maximum static friction force between the rubber tires and the concrete track. The maximum static friction force can be calculated using the formula:
fs max = µs * N
Where:
fs max = maximum static friction force
µs = coefficient of static friction (given as 1.00 for the rubber tires and concrete track)
N = normal force
The normal force is equal to the weight of the car, which can be calculated using:
N = m * g
Where:
m = mass of the car (1500 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the known values:
N = (1500 kg) * (9.8 m/s^2)
N = 14,700 N
Now, we can calculate the maximum static friction force:
fs max = (1.00) * (14,700 N)
fs max = 14,700 N
Since the forward force provided by the car's wheels is 1000 N, which is less than the maximum static friction force, the car will not slide. Therefore, there is no specific time at which the car begins to slide in this scenario.
a. F=ma, a=Force/mass, and that acceleration is tangential to the circle at each instant.
b.
force friction=mu*mg
force cnetripetal= mv^2/r
at sliding time, they are equal, and v=a*time or
mu*g=a^2*time^2/r
solve for time