# When to use centripetal vs. radial acceleration

1. Mar 1, 2013

### fredrogers3

My question is more general than anything. When do I use centripetal acceleration vs. radial acceleration. The solutions in my physics book define a in polar coordinates as positive (v^2)/r. However, my professor uses -((v^2)/r). When do I know when to use each respective form?
Thanks

Last edited: Mar 1, 2013
2. Mar 1, 2013

### voko

What you have in both cases, mv^2/r and -mv^2/r, is not acceleration. It is a force. By Newton's third law, a force is always paired with a counteracting force. When something is made to go round, it experiences a force with magnitude mv^2/r; but it equally exerts a force with magnitude mv^2/r onto whatever causes it to go round - in the opposite direction. It is up to you to assign signs to these two forces, but they will have to be opposite.

3. Mar 1, 2013

### haruspex

-v2/r makes the directions of measurement of the radius and of the acceleration consistent.

4. Mar 2, 2013

### ehild

Speaking about centripetal acceleration, we mean acceleration towards the centre and magnitude v2/R.
In polar coordinates, the acceleration vector has a radial and an angular component. The radial unit vector points outward, so the radial component of acceleration along a circle is negative, -v2/R.

If you are in a rotating frame of reference, you experience an inertial force because of rotation, called "centrifugal force" mv2/R, which tends to push a free object outward with acceleration v2/R.

ehild

5. Mar 2, 2013

### rcgldr

Centripetal acceleration is the acceleration related to -v^2 / r. Radial acceleration in polar coordinates would be an additional component of acceleration. For example, imagine a puck sliding on a frictionless surface attached to a string that goes through a hole in the frictionless surface. If string is not being pulled inwards or released outwards, then the path of the puck is circular, and the tension in the string results in centripetal acceleration of the puck. If the string is pulled inwards or released outwards, there's an addtional radial component of acceleration, and the path of the puck would not be a circle (it could be a spiral, ellipse, parabola, hyperbola, ... ).

6. Mar 2, 2013

### fredrogers3

I appreciate all of the help. Maybe I will better understand this by asking this:

A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about a vertical axis at frequency f. Determine the angle at which the bead is in equilibrium.
http://www.webassign.net/gianpse3/5-45.gif

When I draw my free body diagram and write Newton's relations, I am using negative (v^2)/r for acceleration. Both of my normal force components come out positive. However, using the negative acceleration above yields the incorrect answer. I have an extra negative in the answer. The book uses the positive acceleration. Does this have to do with the body rotating?

7. Mar 2, 2013

### voko

I think you should show your attempt.

8. Mar 2, 2013

### fredrogers3

Along the y-direction, -Fg+Fncosθ=ma, a=0
Along the r-direction, Fnsinθ= -m((v^2)/r)

Once again, the book has the above as positive. I get the final answer correctly, with the exception of a negative.

9. Mar 2, 2013

### rcgldr

y-direction and r-direction is not polar coordinates. You should be able to seperate the forces into x and y directions. In this case, positive and negative will depend on the directions of the x and y components of force exerted onto the bead at some instant in time.

The bead is technically never in equlibrium, since the bead experiences a net horizontal force resulting in centripetal acceleration. There is an angle that corresponds to no net vertical force on the bead.

Last edited: Mar 2, 2013
10. Mar 2, 2013

### ehild

The bead is in equilibrium with respect to the rotating hoop. From a rest frame of reference, it is moving along a horizontal circle.

The drawing on the left shows the forces seen from the inertial frame of reference. Gravity and normal force act on the bead, and the resultant force is equal to the centripetal force, pointing inward.

The drawing on the right shows the bead from the rotating frame of reference. The bead is in rest in that frame of reference, so the resultant force is zero.
The forces are gravity and normal force, and also the virtual centrifugal force, that points outward.

ehild

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11. Mar 3, 2013

### voko

There are a couple of problems in these equations. The most obvious one is that the "r-direction", if we follow the notation in the picture, is not orthogonal with the "y-direction", so there is a component of the weight acting in that direction, too. Plus, the component of the normal force in that direction would be just Fn, not Fn sin θ.

The basic equation is ma = Fn + Fg. In the y-direction: 0 = Fny + Fgy => Fny = - Fgy = mg.
In the x-direction: max = Fnx + Fgx = Fnx.

The acceleration in the x-direction is ax = -ω2d, where d is the distance from the axis of rotation. Note the sign is negative, because the acceleration is always opposite to the distance. As can be seen from the picture, d = r sin θ, so ax = -ω2 r sin θ.

The tricky part is in computing Fnx. It seems it should be Fnx = Fn sin θ. This, however, leads to an unexpected result in the end. Angles can be positive and negative, and the usual convention is that the positive direction is counter-clockwise (thus the angle shown in the picture is in fact negative). Following the usual convention, it can be seen from the picture that for a positive angle, the x-component of Fn should be negative, and positive otherwise. Thus Fnx = -Fn sin θ and the equation in the x-direction is

-m ω2r sin θ = -Fn sin θ.