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I can realize that they are different, but can't understand what actually creates a rotational motion. Rotational motion doesn't mean just moving in a circle. There must be rotation of the position vector for rotational motion. For that, torque is required. Centripetal force,on the other hand,rotates the velocity vector. If this is the difference,what actually makes a door to rotate? Most book writes, it is torque, since the postition vector of the door's edge is rotating with respect to a hinge. But isn't there centripetal force,which is changing the direction of the velocity vector of the edge of the door to let it move in a circle?? What is actually happening? What is the difference between these two (apart from units!)??

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You need a torque/force to start or stop a door, because an external torque makes something slow down how fast it rotates or speed up how fast it rotates or change the direction of how it rotates.

But an external torque is not why the door keeps rotating. Imagine that you and your friend are out of the ice and you grasp each other hands facing each other. If a torque gets you spinning (say someone/something pushes your friend east and pushes you west) then you have hold on to each other to rotate in that symmetric non-expanding way. This is how rigid objects rotate, they are holding themselves together because they are stretching like a rubber band.

So let's now look at a top. At rest it is a certain size, the size that is a natural equilibrium based on the stuff it is made of of, how far the atoms can be and hold each other in place. But if you spin it, then it gets fatter in the middle, but that equilibrium position resists you making it bigger so it tries to pull itself back together, but the direction it pulls itself together is orthogonal to the velocity, so it doesn't change the speed, so it just keeps spinning. This force is a force that the object is exerting on itself. The parts on the outside are pulling the inner parts in an outward direction while those inner parts are pulling the outer parts in an inwards direction, but nobody goes any faster or slower because the directions they are pulling are orthogonal to the velocity.

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Centripetal force and torque are two very different things. Applying a torque to the door will cause angular acceleration while the centripetal force preserves rotation. The centripetal force is required to maintain circular motion and is provided by the bonds between the molecules in the door. A door is fairly rigid and so this force can be ignored as the "right amount" of centripetal force will be provided.

Your confusion is between torque and centripetal force. The torque produces the angular acceleration which changes the door's angular displacement, which causes the door to spin. The centripetal force, which is provided by the bonds between the molecules and in the hinge, keeps the door in its circular motion (otherwise the door would fly in a straight line).

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First, let's grasp the idea that there is no force that is "the centripetal force." There are forces like gravity, Lorentz (EM), tension, and spring which can have components directed toward a center of curvature. A net force toward a center of curvature will have an associated centripetal acceleration, and the product of mass times this centripetal acceleration is commonly (and IMHO, confusingly) called centripetal force. But $m\frac{v^2}{r}$, while having force units, is not a force. It is mass ($m$) times acceleration ($\frac{v^2}{r}$).

The torque produced by a force involves the force component acting perpendicular to the line from a mass point to some defined point, usually a point on a fixed axis (but not always). This perpendicular component times the distance from the defined point to the application of the force is a torque. See the contrast between the perpendicular component used in the torque and it's product with the distance, and the radial (and usually) parallel component of force (and no distance multiplier) which contributes to the centripetal acceleration. Centripetally-directed force components product zero torque about the center of curvature.

Let's use a door swinging on hinges about a vertical axis to contrast the two. A net torque about the vertical axis colinear with the hinges will produce an angular acceleration about the hinges. This torque might be produced by someone's hand exerting a horizontal force perpendicular to the door.

As the door rotates, every part of the door has a changing velocity, changing in direction. This means (by Newton's First Law) that there is some force acting perpendicular to the velocity, and therefore, parallel to the door. This force is the electrostatic force holding the door together, and it acts toward the axis of rotation. It is producing a centripetal acceleration. It keeps the particles of the door from moving on straight lines. It didn't produce the rotation, but it holds the pieces of the door in a circular path.

Rotation is also a matter of perspective. Consider an airplane flying over a highway at constant speed and constant height. Let's ignore the slight curvature of the Earth for a moment. One might say the airplane is flying in a straight line, and that, to first order, is true. Now imagine a person sitting beside that highway, watching the plane through binoculars. That person will have to rotate the direction of the binoculars in order to watch the plane. One could then say the airplane is rotating around the person, which is also true.

Torques are always calculated about points, but there is no absolute point one must use. There are, however, some points which are more convenient to use. Centripetal accelerations, however, are always toward the instantaneous center of curvature. For circular paths, that happens to be the center of the circle.

Torques change angular momentum vectors and result in rotations of massive particles about points. Forces change momentum vectors and result in accelerations of massive objects. The rotation of a vector does not require a torque because the vector itself is not massive.

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  • $\begingroup$ I didn't thank you for your answer when you posted at the mid-year. Sorry for that. Didn't look back again at the answer. However, +1 lately:) $\endgroup$ – user36790 Dec 21 '15 at 16:03