Refresh my memory on centripital force/acceleration

In summary, the conversation discusses drawing a free body diagram for a ball spinning in a vertical circular path and determining its minimum and maximum velocity without leaving the circular path or breaking the string it is attached to. The net force equation at the top would include gravity and another force pulling down, while at the bottom it would include gravity and the upward tension force. The net force equation would be Fnet = mv²/r = mg + T at the top and Fnet = mv²/r = T - mg at the bottom.
  • #1
benji
48
0
Alright so we did this stuff at the beginning of the year and haven't touched it until now and I can't remember exactly how it works...

Say a ball is being spun in a vertical circular path and you need to draw a FBD for it at the highest point and lowest point on the circle.

At the top you would just have the force of gravity pulling down, correct? Or would you have the centripital force pulling the ball downward towards the center of the circle too?

At the bottom you would have the force of gravity pulling down and the centripital force pulling the ball towards the center of the circle, correct?

Hrm... But this is screwy in my head.

Because now I need to figure the minimum velocity the ball could have without leaving the circular path at the top of the circle... So my net force equation would look like: (damn I wish I knew how to do all the fancy markup on the forum)

Fnet=Fg=ma

So my minimum velocity would end up being v=sqrt(gR)?

Then at the bottom what would my net force equation look like if the maximum tension the string could have without breaking was Tmax and I needed to derive an equation for Vmax (the maximum speed the ball can have without breaking the string).

I know I'm wrong with something here...
 
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  • #2
At the top you would just have the force of gravity pulling down, correct? Or would you have the centripital force pulling the ball downward towards the center of the circle too?

The centripetal force is the NET force on the object (i.e. the vector sum of all the real forces acting).

At the top you will have gravity pulling down and some other force also pulling down (e.g. the tension in the string the ball is attached to). The NET force is downwards (centripetal).

At the bottom you would have the force of gravity pulling down and the centripital force pulling the ball towards the center of the circle, correct?

The tension would pull up, and gravity would pull down. The sum of the two forces gives a NET centripetal force upwards.

So my net force equation would look like: (damn I wish I knew how to do all the fancy markup on the forum)

Fnet=Fg=ma

Your net force equation at the top, for example, would look like this:

[tex]F_{net} = \frac{mv^2}{r} = mg + T[/tex]

(taking the downward direction as positive).

You can do the net force at the bottom.
 
  • #3
Thanks James, really appreciate it.
 

Related to Refresh my memory on centripital force/acceleration

1. What is centripetal force and how does it relate to acceleration?

Centripetal force is a force that acts towards the center of a circular path, keeping an object in uniform circular motion. It is directly related to acceleration, as the force is responsible for continuously changing the direction of the object's velocity, resulting in a change in acceleration.

2. How is centripetal force different from centrifugal force?

Centripetal force is a real force that acts on an object, pulling it towards the center of a circular path. Centrifugal force, on the other hand, is a fictitious force that appears to push an object away from the center of a circular path due to its inertia.

3. What are some examples of centripetal force in everyday life?

Some examples of centripetal force include the force of gravity that keeps the moon in orbit around the Earth, the tension force in a string that keeps a yo-yo moving in a circular motion, and the force of air resistance on a car tire as it goes around a curve.

4. How is centripetal force calculated?

The magnitude of centripetal force can be calculated using the formula F = m * v^2/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

5. Can centripetal force be increased or decreased?

Yes, centripetal force can be increased or decreased by changing the speed or radius of the circular motion. Increasing the speed or decreasing the radius will result in a greater centripetal force, while decreasing the speed or increasing the radius will result in a smaller centripetal force.

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