A 1900kg car starts from rest and drives around a flat 56-m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1000N.
a. What is the magnitude of the car's acceleration at t=13s?
b. What is the direction of the car's acceleration at t=13s? 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?
Attempt at a solution:
a= sqrt(tangential acc. sq + radial acc. sq)
a(t)= F/m
a(r)= mv^2/r
here v= at (?)
But I got a huge acceleration for doing that and it's a wrong answer. Where did i go wrong?
Thanks in advance.
I am wondering what forward friction is. It is hard to assume it is zero, but that affects what the forward acceleration.
1000-frictionforward= mass*acc forward
otherwise, what you laid out is the correct procedure.
a= sqrt((netforce/mass)^2+ (at)^2/r)
Notice I disagree with your a(r), I don't see the need for mass to be there.
Thank you so so much. Well I got the first two answers correct now. But in c i think i should use friction force which should be rubber upon concrete? Co-efficient of kinetic friction for that is 0.8. (I need the kinetic friction co-efficient here right?) But i'm not sure how to do this part. What will the approach be?
You need the coefficient
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To solve this problem, we need to consider the forces acting on the car and apply Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration (F = ma).
a. To find the magnitude of the car's acceleration at t=13s, we can use the equation F = ma. The forward force provided by the car's drive wheels is constant at 1000N. Therefore, the net force acting on the car is also 1000N since there are no other forces mentioned in the problem.
F = ma
1000N = (1900kg) * a
Solving for 'a', we find that a = 1000N / 1900kg ≈ 0.526 m/s².
So, the magnitude of the car's acceleration at t=13s is approximately 0.526 m/s².
b. To find the direction of the car's acceleration at t=13s as an angle from the r-axis, we need to consider the circular track and the forces involved. The acceleration of an object moving in a circular path is always directed towards the center of the circle, which corresponds to the radius. Hence, the acceleration in this case is radial and points towards the center of the circular track.
To determine the direction as an angle from the r-axis, we can use the concept of the tangent function.
tan(θ) = a(r) / a(t)
tan(θ) = (mv²/r) / (at)
We know the car's mass (m = 1900kg), the radius of the circular track (r = 56m), and the magnitude of the car's acceleration (a = 0.526 m/s²). We're trying to find θ.
First, we can find the tangential acceleration using the formula:
a(t) = v / t
However, since the problem does not provide the car's initial velocity or information to calculate it, we cannot determine the tangential acceleration or the angle θ accurately.
c. To determine the time at which the car begins to slide out of the circle, we need to consider the forces involved when the car is on the verge of sliding.
The maximum frictional force (f_max) can be calculated using:
f_max = μ_s * N
where μ_s is the coefficient of static friction and N is the normal force.
For the car not to slide, the maximum frictional force must balance the maximum centripetal force acting towards the center of the circle. Therefore, we can equate these two forces:
mv²/r = μ_s * N
We need to find the velocity (v) at which the car will slide out of the circle. At this point, the frictional force will be at its maximum value.
But without knowing the coefficient of static friction (μ_s) between the rubber tires and the concrete track or any other relevant information, we cannot determine the time at which the car begins to slide out of the circle.