Centripetal/Tangential Acceleration

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The problem involves a pebble in a tire tread that is held by static friction as the car accelerates. The maximum static friction force is 3.6 N, and the calculation for the speed at which the pebble dislodges was attempted using centripetal force equations. The initial calculation of 10.677 m/s was deemed incorrect due to an error in mass conversion. It was clarified that centripetal force is not an applied force but rather a necessary component of the net force to maintain circular motion, and the gravitational force acting on the pebble is negligible. Correcting the mass conversion is essential for finding the accurate speed at which the pebble flies out.
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Homework Statement



A 1.2 g pebble is stuck in a tread of a .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?


Homework Equations



net F= (mv^2)/r
t=I*α
v=ω*r

The Attempt at a Solution


I am self-studying physics right now and encountered this puzzling problem. Because it was in the rolling motions, I immediately tried to solve it using torque. Unfortunately this didn't result in anything, due to absence of distance or time. The only rotational force that dealt with velocity was centripetal, so I considered the force as centripetal force. Using the equation F=mv^2/r, I had (3.6N)=(.012kg)(v^2)/.38m

And finally, v=sqrt((.38m*3.6N)/.012kg)=10.677 m/s...

The textbook says this is wrong. Thus, my question is two-tiered: first, what did I do wrong? second, it looks like this is centripetal force, but why? (Isn't centripetal force inward, thus making friction face outward?)
 
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PhysicsVA said:

Homework Statement



A 1.2 g pebble is stuck in a tread of a .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?


Homework Equations



net F= (mv^2)/r
t=I*α
v=ω*r

The Attempt at a Solution


I am self-studying physics right now and encountered this puzzling problem. Because it was in the rolling motions, I immediately tried to solve it using torque. Unfortunately this didn't result in anything, due to absence of distance or time. The only rotational force that dealt with velocity was centripetal, so I considered the force as centripetal force. Using the equation F=mv^2/r, I had (3.6N)=(.012kg)(v^2)/.38m

And finally, v=sqrt((.38m*3.6N)/.012kg)=10.677 m/s...

The textbook says this is wrong. Thus, my question is two-tiered: first, what did I do wrong? second, it looks like this is centripetal force, but why? (Isn't centripetal force inward, thus making friction face outward?)

Your solution looks fine to me. You've calculated the force needed to keep the stone on the tire at velocity v and equated it to the maximum frictional force.
 
Don't forget that gravity is an additional force acting upon the pebble.
 
PhysicsVA said:
And finally, v=sqrt((.38m*3.6N)/.012kg)=10.677 m/s...
Check your conversion of the pebble mass.
Isn't centripetal force inward, thus making friction face outward?
Centripetal force is not an applied force. It is the radial component that the net force must have in order to keep the object moving at constant radius. The pebble will be dislodged when the static friction is at its maximum value, yet the radial component of the net force is less than the centripetal force required.
As rgcldr posted, the net force comes from adding the static friction and the gravity on the pebble. However, the gravitational force will be so small that it can be neglected here.
 
haruspex said:
Check your conversion of the pebble mass.

Whoa. Thank you. I missed the conversion.
 
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