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Homework Help: Torque and radius. a confusion!

  1. Feb 6, 2008 #1
    why do we define the torque as FxR?(both in vectors) why does the same force applied to a body( say, a torque wrench) produce a smaller acceleration if applied at a point closer to the center of rotation?
  2. jcsd
  3. Feb 6, 2008 #2
    Well, FxR means the magnitude of F times the magnitude of R times the sin of the angle between them. So if a force is applied at a shorter radius from the axis of rotation, by definition the magnitude of R will be smaller and thus produce a smaller torque.

    So, the magnitude of the torque (a force applied at a distance) is smaller, and therefore the acceleration produced by this force must also be smaller.
  4. Feb 6, 2008 #3
    well, that is only the magnitude of torque!?, but the x cross product is far more than F.R.Sin(a)

    FxR is cross product, my advice is to look at here:


    why?!, well, we must give a meaning to the rotational force the same way me give to angular velocity and angular aceleration..

    why do you say that? are you talking about angular aceleration?
  5. Feb 6, 2008 #4
    His second question asked why something was 'larger'; this implies magnitude. I don't mean to say that the cross product is only these things, but the magnitude of the torque (pertinent in explaining when it is larger or smaller) is indeed the magnitude of the force times the magnitude of the radius times the sin of the angle between them.
  6. Feb 7, 2008 #5
    @ all
    i know the formula of torque. but in case of translational motion, the same force causes the same acceleration in bodies of same masses. but in case of rotational motion, the same force causes different angular accelerations in bodies of same moments of inertia depending on the distance of application from the rotational axis. (assuming same angle theta). can you explain this phenomenon in a more understandable way other than giving the formula? i'm not able to comprehend it.
  7. Feb 7, 2008 #6
    My apologies.

    Since we know the formula to be FRsin(theta), and since we know the angle and the force are the same, we know that R is the only thing left to affect the torque, by definition.

    We know that R is the radius - that is, the distance from the axis of rotation at which the force is applied. So, if you're holding a wrench and trying to tighten a bolt, you're more likely to put your hand at the end of the wrench (that is, as far as possible from the bolt to be tightened). Why is this? For one thing, it's rather intuitive and quickly discovered after handling the tool for a very short time. If you try to tighten the bolt with your fingers, the distance from the center of the bolt is very small and would thus require a very large force in order to produce the same torque.

    Alternatively, picture a teeter totter made of a bench that is not centered on the fulcrum. If one side has twice the length as the other (that is, if its farthest distance from the fulcrum is twice the other), this corresponds to the longer side having 2R and the shorter side having R as the distance from the axis of rotation (the fulcrum). So, in order to produce the same torque, the side of length 2R needs just half the force (assuming the same angle) as the side with length R.

    I'm sorry if I've been answering the wrong question (and I suspect I have). If you're asking why all the above is true, my answer would be that it simply works out to be this way. The universe is what it is and we just happened to find a reasonably accurate mathematical model that allows us to describe it quantitatively. Other than making intuitional sense, I'm not sure how much of an explanation actually exists.
  8. Feb 7, 2008 #7
    I'm not exactly sure of how exact this definition is but, when a force acts on an object, its generally simple to picture translational motion in terms of mechanics because it consists of F = ma, in which all parts of the object move with a translational acceleration a = F/m. However, objects also experience rotational motion and harmonic (vibrational) motion as well.
    The reason why the magnitude of acceleration is larger towards the edge of a rotating object rather than the inside is simply because each part of the object is rotating a length 2*pi*r, r being the distance of that point from the axis of rotation. Each part of the object travels one full period (circle) at the same time, however, points toward the edge of the object have to travel a greater distance (due to the larger distance from the rotation axis) and they travel this distance in the same amount of time (so, their acceleration would be larger).
    If a spherical object were slipped on to a frictionless axle that runs through its center, it would stay at rest on the axle unless a force acts on it. If say you directed a force on the sphere and caused it to rotate, not all parts of the object experience the same tangential acceleration. Therefore, rotational motion isn't simply analyzed using F = ma. The concept of torque is to simply measure how effectful a force is in causing an object to rotate. The way to measure it is to measure the distribution of the force along the radius of the object (F x r). It is a cross product vector and points perpendicular to the plane of rotation.
    Last edited: Feb 7, 2008
  9. Feb 9, 2008 #8
    wikipedia says torque is the easiest way to explain mechanical advantage. so when you've greater torque for a larger R vector, wat do you lose? (like requiring a higher distance of force application in case of compound pulleys)
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