Mastering Physics: Rolling Motion, Pebble stuck in a Tire by Static Fr

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SUMMARY

The discussion centers on calculating the speed at which a 1.2 g pebble, stuck in a 0.76 m diameter tire, will fly out due to static friction. The maximum static friction force is 3.6 N. The correct approach involves using the centripetal force equation, Fc = mv²/r, leading to a calculated speed of approximately 33.76 m/s when the pebble dislodges, rather than the initially mentioned 17 m/s. The solution emphasizes the importance of not dividing the final velocity by 2, as the speed at the top of the wheel is not halved.

PREREQUISITES
  • Understanding of centripetal force and its application in rotational motion
  • Familiarity with the equations of motion for circular dynamics
  • Knowledge of static friction and its role in maintaining motion
  • Basic skills in algebra for rearranging equations and solving for variables
NEXT STEPS
  • Study the derivation and application of the centripetal force equation, Fc = mv²/r
  • Explore the relationship between linear and angular velocity, specifically v = ωR
  • Investigate the effects of static friction in different scenarios involving circular motion
  • Learn about the moment of inertia and its significance in rotational dynamics, particularly I = 1/2 * M * R²
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators seeking to clarify concepts related to centripetal force and static friction.

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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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Problem Solved!

I found the solution, I used the centripetal force equation like I explained above, but I did not divide by 2 at the end.

Fc=mv2/r
v=√(Fcr/m)
v=√[(3.6 N)(0.38 m)/(0.0012kg)]
v=33.76388603 m/s
v=38 m/s
 

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