Angular motion problem (wheels on a car coming to a stop)

In summary, the driver of a car traveling at 12.0 m/s with tires of radius 0.40 m applies the brakes and undergoes a constant deceleration of 1.90 m/s^2. The car comes to a stop in 37.89 meters, and the number of revolutions made by each tire before stopping is 15.076 revolutions. However, this answer may not be correct due to possible errors in calculation or input.
  • #1
xregina12
27
0
The driver of a car traveling at 12.0 m/s applies the brakes and undergoes a constnant decelertion of 1.90m/s^2.
How many revolutions does each tire make before the car comes to a stop assuming that the car does not skid and that the tires of radii of 0.40 m? answer in units of rev.

I used the equation Vf^2=Vi^2+2ad
0=144+2(-1.90)(d)
d=37.89meters
d=r(θ)
θ=37.89/0.40=94.725 radians
revolutions=94.725/(2pi)=15.076 revolutions.
However, I did not get a correct answer. Can anyone help? Did I do this the right way?
 
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  • #2
You've worked out the distance it takes to stop correctly, but I don't follow your working after that. You know the radius of the tyre is 0.4m, so it's circumference = 2Pi*0.4=2.5137. The car stops in a distance of 37.89 meters, so the number of revolutions is just the distance it takes to stop divided by the circumference of the tyre. Hope that helps.
EDIT: Actually looking at that method I get the same answer as you, so I'm not sure why that's not the correct answer. If this is from an online homework site are you entering the correct number of decimals?
 
Last edited:
  • #3


Your approach is correct, but there are a few things to consider. First, make sure you are using the correct units for each variable in the equation. In this case, the initial velocity (Vi) should be in meters per second (m/s), the acceleration (a) should be in meters per second squared (m/s^2), and the distance (d) should be in meters (m). You may have made a mistake in converting the initial velocity from 12.0 m/s to 144 m/s.

Also, make sure you are using the correct formula for angular motion. The formula you used, Vf^2=Vi^2+2ad, is for linear motion. For angular motion, the formula is ωf^2=ωi^2+2αθ, where ω is the angular velocity (in radians per second), α is the angular acceleration (in radians per second squared), and θ is the angular displacement (in radians). In this case, ωf=0 (since the car comes to a stop), ωi=12.0 m/s / 0.40 m = 30 radians/s, and α=-1.90 m/s^2 / 0.40 m = -4.75 radians/s^2. Plugging these values into the formula, we get:

0=30^2+2(-4.75)(θ)
θ=30^2/9.5=94.74 radians

Finally, to convert radians to revolutions, we use the formula θ=2πn, where n is the number of revolutions. So, n=94.74/(2π)≈15.08 revolutions.

Therefore, the correct answer is 15.08 revolutions. It's possible that the discrepancy in your answer was due to rounding errors or using the wrong formula. Make sure to double-check your calculations and units to get an accurate answer.
 

What is angular motion?

Angular motion is the motion of an object around a fixed point or axis, such as a wheel rotating on its axle.

How is angular motion related to a car coming to a stop?

When a car is coming to a stop, the wheels are rotating around the axle, and this motion is an example of angular motion.

What factors affect the angular motion of the wheels on a car?

The angular motion of the wheels on a car can be affected by factors such as the force applied to the wheels, the mass of the car, and the friction between the wheels and the ground.

How does angular motion affect the braking distance of a car?

The angular motion of the wheels affects the braking distance of a car because it determines how quickly the wheels will stop rotating, which in turn affects how quickly the car will come to a stop.

How do scientists study and solve angular motion problems?

Scientists use mathematical equations and principles, such as Newton's laws of motion, to study and solve angular motion problems. They also use tools such as force sensors and motion detectors to collect data and analyze the motion of objects.

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