The mysterious centripetal force

By Marcus Wilson 14/03/2011 10

Last week I came face to face with another physics misconception with some of my students. I do think that, as I get more experienced teaching, I’m getting better at picking up on where students are having problems. But it’s a very difficult thing to do.

Last week it was circular motion The students were looking at a fairly simple problem (well, I thought so, they are third year mechanical engineering students after all). In a nutshell, it’s this: Imagine you have a T-shirt in a spin dryer (top loading). If you know the coefficient of friction between the T-shirt and the wall of the drum, and the drum’s radius, you should be able to work out the minimum rotation speed for the T-shirt to stick to the wall of the drum, rather than slide down to the bottom. So, what is it?

As with all force problems, a good way to start is by drawing a free-body diagram, where we put in arrows for every force acting on the body. The total of these arrows (added as vectors), or the net force, is just the mass times acceleration, by Newton’s 2nd law. So, I put a sketch on the board and asked the students to tell me what forces are acting on the T-shirt.  There is of course gravity on the T-shirt. That’s balanced by the frictional force of the wall on the shirt (otherwise it would slide down.) Then there’s the inward force exerted by the reaction of the wall on the T-shirt. Now, that’s all the forces there are. I could have just carried on with the analysis,  but instead I asked the class whether that was it  "Have we got all the forces?"  I expected a ‘yes’ answer, but a student gave me the response "There’s the centripetal force as well." 

There’s a clearly a misconception about circular motion buried in that comment.   Yes, there is centripetal force, but it’s the reaction of the wall on the T-shirt that provides this force. We’ve got it already. The centripetal force isn’t an ‘extra’ force – that would be counting it twice. This is a tricky concept to present clearly. The best I can manage now is that it’s the fact that there exists a force towards the centre of rotation that CAUSES the object to move in the circle, not that the fact that an object moves in a circle causes it to have a centripetal force. 

Centripetal force is not a ‘magic’ force, that mysteriously appears when you turn corners. There is always a very physical force that provides the centripetal force, whether it’s the tension in a string (for whirling a mass-on-a-string) around, friction from the road on your car tyres (for driving a car around a circle), the lift force from an aeroplane’s wings (for an aeroplane banking) or gravitational attraction (for the earth orbiting the sun).

I’m not sure I did a great job during class of explaining this, but at least I know now to look out for this misunderstanding with students in first year. And it also demonstrates that if you don’t ask questions of the students, you really don’t have a clue what they are thinking.

10 Responses to “The mysterious centripetal force”

  • I do a lot of cycling, which gives plenty of time for thinking…

    One thing that always puzzled me a bit: I can be rolling along, lean into a turn, and come out of a 90º turn with (almost) all the speed I had going in, despite the fact that I haven’t peddled, and despite the fact that I’ve lost all my original momentum and gained a whole new batch (since the vectors are perpendicular, right?).

    Of course, there is gravity… so I tried to think of a simpler example. This is what I came up with:

    Imagine a ramp. The ramp transcribes a quarter-circle — it starts off horizontal and finishes vertical. At the bottom of the ramp, place a gun. This gun fires flat pellets — something like a coin. The gun is pointed at the ramp.

    So, when you fire the gun, a pellet comes out with some initial horizontal velocity. It then slides along the ramp and ends up vertical. So some force is acting on the pellet towards the centre of the circle. It’s not gravity, which is pointing the wrong way. What is it?
    (if possible, can I specify that the ramp is frictionless?)

  • Marcus said…
    Imagine you have a T-shirt in a spin dryer (top loading). If you know the coefficient of friction between the T-shirt and the wall of the drum, and the drum’s radius, you should be able to work out the minimum rotation speed for the T-shirt to stick to the wall of the drum, rather than slide down to the bottom.

    As with all force problems, a good way to start is by drawing a free-body diagram, where we put in arrows for every force acting on the body.

    I think I recalled a similar problem described in the text-book I used when I was an undergrad student, title “Engineering Mechanics – Statics” by Meria & Kraige. The object wasn’t a t-shirt but a small cylindrical rod. I might dig this book from my old collection in the shed.

    Do you use this book as a current text-book Marcus? I know that the School of Engineering in Auckland is still prescribing it as a text book for those who are doing mechanical engineering. It was an excellent book in static analysis via free-body diagrams.

  • I am on the equator. I weigh 80kg on my bathroom scales. I speed up the earth’s rotation. (I can do that cos I have powers). As I speed up I notice my weight on the scales gets less. it eventually reaches zero and i feel myself just floating. Gravity accelerates (pulls) me towards the centre of the earth. Centripetal force is supposed to accelerate me towards the centre of the earth.

    What gives Batman?

  • Demonstrating that my velocity would cause me to disappear in a straight line at a tangent if the magical centripetal force suddenly disappears seems to offer an answer……

  • Repton, I think that the problem you’re describing can be derived using rotational energy conservation, since the pellet is traversing a surface of a frictionless quarter circle of say radius R :

    The rotational intertia (I) of the pellet of mass (m) on the frictionless quarter circle of radius (R) is:

    I = m*R^2

    Conservation is then:

    Rotational kinetic energy at the bottom = Potential Engergy + Rotational kinetic energy (at the top at height R )

    (I*Wi^2)/2 = m*g*R + (I*Wf^2)/2

    where :

    Wi – initial angular velocity (at the bottom)
    Wf – final angular velocity (at height R at the top)

    But angular velocity W = V/R (ie, velocity V divided by radius R), so Wi = Vi/R and Wf = Vf/R. Both Vi and Vf are initial & final velocity.

    After substitutions of variables into the energy conservation relation above, you then come up with:

    Vf = sqrt(Vi^2 – 2*g*R) , where g is gravitational constant.

    Vf is the pellet when it reaches the height of radius R which is the end of the frictionless quarter circle surface that it slided up on. The force is still centripetal. The pellet is accelerated towards to the center of the quarter circle (where radius R is measured from) and this is no different to a pendulum with a string of length R tied to a pellet of mass m that pushes at the bottom with initial velocity Vi and when it reaches the height R (end of the quarter circle) it achieves a final velocity Vf.

  • Ross, I think your weight would not change (since the universal gravitational constant, distance from the centre of mass of the earth, and neither masses change). What would happen if the earth increased it’s speed would be that you would gradually be left behind as the earth spun underneath you.

    Your weight force does two things:
    1. Helps provide friction to keep you stuck to the surface of the earth (Normal force, co-efficient of static friction, etc)
    2. Provides a minor contribution to the centripetal force that turns both you and the earth (the remaining centripetal force comes from the gravitational force from the remaining mass of everything else that is rotating about itself)

    Fun thought experiments!

    • Wow, judging from the comments circular motion obviously is a hard topic.

      Some thoughts…
      Going back to Repton’s comment, when something goes round in a circle (a bicycle turning a corner) of a pellet going round a curved track, (or you going round the centre of the earth – I’ll come back to that) ask yourself what it is that causes you to go in a circle. Remember Newton’s first law, if there ain’t no force, your velocity don’t do nuffink. (No force means the particle continues at the same velocity). So, if something changes direction (which is a change in velocity) there MUST be a force causing it. So, for repton’s bicycle, it’s the friction between the tyres and the ground. If you don’t believe me, try cycling on ice, or in outer space, and see if you can turn corners. The force will act towards the centre of the circle (if you’re turning in a pure circle with your speed not changing – NB speed isn’t the same as a velocity – a velocity has a direction, speed doesn’t) and is called the centripetal force (‘centre-seeking’ force).

      So in Repton’s ramp example, the thing that provides the centripetal force is the very thing that causes it to turn in a circle, namely the reaction force of the ramp on the pellet.

      In the rotating earth example, since you move in a circle, there will be a force on you towards the centre of the circle. What is it? Gravity. If the John’s special powers turned the gravity of the earth off, there would be nothing to cause you to carry out a circular path (except if you glued your feet to the earth or clung onto something) and you’d drift off. So I’d dispute some of the technical content of Sam’s comment – first it’s not friction that holds you in place (you don’t drift off the earth if you are on an ice rink, where friction is pretty low), and second, your weight (which is the gravitational force on you) doesn’t provide a minor contribution to centripetal force – it would be all of it.

      (Physics nerds can calculate the size of the centripetal acceleration at the equator from angular velocity of earth (2 pi radians in 86400 seconds) squared, times radius of earth (6400 km) and you’ll get about 0.034 metres per second squared. Gravity is way bigger than this which means it can comfortably hold you in place. )

  • Marcus said…
    NB speed isn’t the same as a velocity

    Correct. In my example above, I made no differentiation, simply for ease of discussion, where others can follow. I could have labelled that velocity Vi is really Vi*ex which is pointing to the right (where ex is the unit directional vector in the x direction, i.e., from left to right). Velocity Vf is really Vf*ez which is pointing upward (where ez is the unit directional vector in the z direction, i.e., from bottom to top).

    In advanced mechanics & kinematics, those physical observables (linear/rotational momentum, velocity, displacement in 3D) are represented as vectors and the usual rule of vector operation applies, i.e., the dot-product & cross-product, (non) commutative operations, etc,…

    The advantage of using vector notations is, that the physics is more generalizable. In quantum mechanics & atomic physics, representing the state of particles in vector notations and also their operators (angular momentum operator, energy operator), which are complex number based differential (calculus) operators can lead to new solutions when they’re being solved. If they’re not being vectorized, such solutions may not be obvious (or perhaps difficult to solve), when those physical observables are represented as scalar functions (e.g., speed).

    Vectorization lead to solutions which quantify electron’s spin & orbital magnetic dipole moments, atomic shells/sub-shells electron configuration/model, nuclear shell model, spin-orbit coupling, nuclear coupling, atomic selection rule (in chemistry) , and so forth. Raman spectroscopy & NMR (nuclear magnetic resonance) are examples of technology which are designed based on those models/quantities listed above. Of course there are many more applications, but those mentioned are familiar to the public.

    I can’t imagine quantum mechanics being successful without vectorization (via differential operators) of physical observables. Well, establishing the electron configuration & atomic shell model is impossible to derive via scalar functions (speed) rather than vectors (velocity). Classical centripetal force concept is thus fundamental in quantum mechanics, however physical observables are represented as vectors and their operators are also represented in (complex number) vector differential form.

  • I wonder if you could take some time to delve into the following two concepts concerning centripetal force and entanglements.

    First, does centripetal force affect gravitational effect. While I fully understand that the mass of an object doesn’t change and thus the gravitational pull on an object doesn’t change regardless of angular momentum (orbital velocity) would a man standing on a planet with no axial spin “weigh more” (subjectively) without the angular momentum offsetting some of the gravitational pull? (I understand the answer, but I thought it would make for a useful topic of discussion).

    Secondly, I would absolutely love to read an exposition on quantum spin (as in relativistic quantum mechanics) vs. orbital angular momentum. What is the explanation of how or why any quantifiable object has centripetal force rather than the universe spinning around the object? Is this just all gravity pulling in all directions or is there an explanation which would allow for centripetal force to be applied within a Klein space containing only a single set of quantifiable mass that retains the properties observed in classic physics? Basically, does space-time create the basis for Newton’s third law of momentum conservation or does it instead come from a property of entanglement?

    note: I’m not highly educated in physics or mathematics, just smart and learning on my own.