Rotational motion of a car's tires

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SUMMARY

A car accelerates uniformly from rest to a speed of 19.0 m/s in 7.00 seconds, resulting in a distance traveled of 66.5 meters. Given the tire diameter of 64.0 cm, the perimeter of the tire is 0.20106 meters. The tire makes approximately 331.5 revolutions during this motion. The final rotational speed of the tire is 55.25 revolutions per second.

PREREQUISITES
  • Understanding of uniform acceleration in physics
  • Knowledge of circular motion and tire perimeter calculation
  • Ability to convert linear speed to rotational speed
  • Familiarity with basic kinematic equations
NEXT STEPS
  • Calculate the distance traveled using the formula for uniformly accelerated motion
  • Learn how to derive the perimeter of a circle from diameter measurements
  • Explore the relationship between linear speed and rotational speed
  • Study the concept of angular displacement in rotational motion
USEFUL FOR

Physics students, automotive engineers, and anyone interested in understanding the dynamics of rotational motion in vehicles.

ryanl
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A car accelerates uniformly from rest and reaches a speed of 19.0 m/s in 7.00 s. The diameter of a tire is 64.0 cm.

(a) Find the number of revolutions the tire makes during this motion, assuming no slipping.

(b) What is the final rotational speed of a tire in revolutions per second?


How do I set this up?

Thank you.
 
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just think abotu this one for a sec... its not too hard.. kinda basic ... have you tried it yet?
 
Let's play 20 questions. :biggrin:

(a) Can you find the distance that the car travels during its acceleration? Can you find the perimeter of the tire? Can you tell how many times the perimeter fits inside the distance the car travelled?

(b) Do you know what the speed of a certain point on the tire is? Can you find how much time it takes that point to travel all around the tire and return to its original position? Can you now tell how many times it could do so in one second?
 

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