Solve for Angular Displacement with Acceleration: Rolling Motion Help

[wilmerena]

Hi, I am working on the following problem:

As you drive down the road at 17 m/s you press on the gas pedal and speed up with a uniform acceleration of 1.12m/s2 for 0.65s. If the tires on your car have a radius of 33cm , what is their angular displacement during the period of accceleration?

This is what I did so far,

(1.12m/s^2) / 0.65s = 1.72 m/s acceleration

then 17m/s - 1.72 m/s = 15.28

15.28/.33m = 46 displacement?

Im not sure this was right??

 

[Galileo]

(1.12m/s^2) / 0.65s = 1.72 m/s acceleration

?? The acceleration is 1.12 m/s^2 as you stated in the problem.

You can use the formula for the distance traveled when traveling at constant acceleration.

The distance traveled is

d=17(0.65)+1/2(1.12)(0.65)^2=11.3 m

If the wheels don`t slip then the distance traveled when the wheels make one revolution is equal to their circumference 2\pi R

 

[M98Ranger]

Hi there,

You're definitely on the right track! To solve for angular displacement in this problem, we can use the formula:

θ = ω0t + 1/2αt^2

Where:
θ = angular displacement
ω0 = initial angular velocity (in this case, 0 since the tires start at rest)
α = angular acceleration
t = time

First, let's convert the given values to the correct units. The initial velocity and acceleration are both given in meters per second, so we'll need to convert them to radians per second and radians per second squared, respectively.

ω0 = 0 rad/s
α = 1.12 m/s^2 * (1 rad/0.33 m) = 3.39 rad/s^2

Next, we can plug these values into the formula:

θ = 0 * 0.65 + 1/2 * 3.39 * (0.65)^2 = 0.73 radians

Finally, we can convert this back to degrees if needed:

θ = 0.73 radians * (180 degrees/π radians) = 41.8 degrees

So the angular displacement of the tires during the acceleration period is approximately 41.8 degrees. I hope this helps!