Centripetal/Tangential Acceleration

In summary, the question asks for the speed of a car when a 1.2 g pebble flies out of a tread of a .76-m-diameter automobile tire, held in place by static friction of 3.6 N. Using the equation F=mv^2/r, the solution is found to be 10.677 m/s, but the textbook states this is incorrect. After further analysis, it is found that the conversion of the pebble mass was originally missed.
  • #1
PhysicsVA
1
0

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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  • #2
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.
 
  • #3
Don't forget that gravity is an additional force acting upon the pebble.
 
  • #4
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.
 
  • #5
haruspex said:
Check your conversion of the pebble mass.

Whoa. Thank you. I missed the conversion.
 

What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is perpendicular to the object's velocity.

What is tangential acceleration?

Tangential acceleration is the acceleration experienced by an object along the tangent of its circular path. It is a result of changes in the object's speed or direction of motion.

How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

What is the relationship between centripetal and tangential acceleration?

Centripetal and tangential acceleration are perpendicular to each other and together make up the total acceleration of an object moving in a circular path. Centripetal acceleration causes the object to change direction, while tangential acceleration causes it to change speed.

What are some real-life examples of centripetal and tangential acceleration?

Examples of centripetal acceleration include a car turning around a curve, a planet orbiting around a star, and a ball spinning on a string. Tangential acceleration can be seen in a car accelerating or decelerating, a roller coaster going up or down a hill, and a pendulum swinging back and forth.

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