Rolling Motion, Pebble stuck in a Tire by Static Friction

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SUMMARY

The discussion centers on calculating the speed of a car when a pebble, weighing 1.2 g, dislodges from the tread of a 0.76 m diameter tire due to static friction. The maximum static friction force is 3.6 N, leading to a calculated velocity of approximately 17 m/s when the pebble exits the tire. The solution involves using the centripetal force equation and the relationship between the velocity of the tire and the car. The final calculated speed of the car is confirmed to be 17 m/s, aligning with the textbook answer.

PREREQUISITES
  • Understanding of centripetal force and its application in rotational motion
  • Familiarity with the equations of motion, specifically v=(2*π*R)/T and v=ωR
  • Knowledge of static friction and its role in maintaining circular motion
  • Basic principles of rotational inertia, particularly I=1/2*M*R²
NEXT STEPS
  • Study the derivation and application of the centripetal force equation, F=(m*v²)/r
  • Learn about the relationship between linear and angular velocity in rotational systems
  • Explore the concepts of static and kinetic friction in detail
  • Investigate real-world applications of rotational dynamics in automotive engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for practical examples of static friction in action.

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Homework Statement



A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.6 N. The car starts from rest and gradually accelerates on a straight road. How fast is the car moving when the pebble flies out of the tire tread? I know the answer is 17 m/s however I don't know how to get it.

Homework Equations



v=(2*\pi*R)/T
v=ωR
\alpha=τ/I
I=1/2*M*R2
Fc=(m*v2)/2

The Attempt at a Solution



I tried using the centripetal force equation and rearranged for velocity, using the maximum static friction as my force. I then divided that number by 2 to get the speed in the centre of the wheel (since the speed at the top of the wheel is twice the speed of the vehicle).
 
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I believe the approach suggested by the OP gives the correct answer.

F=(mv^2)/r
Rearrange for v...
V=SQRT(Fr/m)
= SQRT(3.6*0.38/0.0012)
= 33.76m/S

This is the Velocity of the top of the wheel. The Velocity of the car is half that..

33.76/2=16.88m/S

Book answer 17m/S.
 

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