Help with rolling motion

  • Thread starter wilmerena
  • Start date
  • #1
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Hi, Im 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?? :frown:
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
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(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

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

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

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