That's correct, provided you don't move the rotation axis of the object, only the point where you push.
but if you consider applying a tangential acceleration at a point 2r from the axis of rotation, and applying the same acceleration at a point r from the axis of rotation, obviously the angular acceleration will be double at radius r, because the object will be sweeping the same distance but at half the radius.. so this suggests the angular acceleration is inversely proportional to the lever arm of the point accelerated.
This is also true, but think carefully about the phrase "applying an acceleration." Exactly how can you apply an acceleration? Answer: you must use a force to produce it. It should be clear that the amount of force required to produce a given tangential acceleration is NOT proportional to the acceleration independent of radius. For example, pushing on the rim of a wheel may cause a tangential acceleration at the point where you push, but how much tangential acceleration will the SAME force produce it you push right on the axis instead? None, because the axis is fixed. Since the size of the force is not locked to the size of the tangential acceleration, it is perfectly possible that angular acceleration may be inversely
proportional to the lever arm for constant tangential acceleration
, and yet at the same time the size of the angular acceleration may be directly
proportional to the lever arm for constant tangential force
I know I am probably missing something really obvious, possibly something to do with the way the torque distributes throughout a body so that it is not the same as accelerating a single point on a rotating body to a certain speed.. but please help if you can explain these things for me! it's been bugging me all day... thanks very much
First, let me say that nothing is really obvious until after you learn it, and then everything is really obvious. So don't beat on yourself too much about it.
Here is a way to look at why torque is proportional to the lever arm, which doesn't mention energy.
Start by thinking about what torque does NOT depend on. Take a rigid object with a fixed axis, pick a point somewhere on it away from the axis, and push there with a force F. Suppose you push at an angle, not straight at the axis. The point you are pushing on is only free to move along a single direction, tangent to a circle around the axis, because you can't move it closer to the axis without compressing the object, and we are assuming it is rigid. Now resolve your force into radial and tangential components. The point cannot move in the radial direction, so that force component will be unable to make it move. The tangential component is pushing in a direction that the point IS free to move it, so it WILL cause motion, and the object will rotate. Conclusion: the torque depends ONLY on the tangential component of the force.
Now comes the tricky part. Suppose you want to think about the effect of pushing at different points on the object. You can analyze it like this: Pick two points A and B on the object, and attach a compressed spring between them. Suppose it makes a force F on A and -F at B. These forces push directly along the line joining A and B. Will the object start to rotate? Of course not, since there is no external force on the system consisting of mass and spring together. Therefore, the net torque exerted on the object by the spring is zero. This means that a force F at A must exert exactly the opposite torque of a force -F at B. Therefore, if you placed a force F (instead of -F) at B, the torque it makes must be the same as F makes at A. Conclusion: If you slide the point of application of a force along a line parallel to the force vector, the torque will not change. Therfore, torque depends only on this line itself, not the exact position along it.
Finally, put these two conclusions together. Suppose point A is at a distance R from the axis, and you push with force F at an angle theta (meaning that theta is the angle between the force vector and the radial line from the axis to A). The tangential component of F is F*sin(theta). Now slide F along a line parallel to itself to another point B, and let B be the point on the line that is NEAREST to the axis. At B, the force is perpendicular to the radius. The distance from the axis to B is Rsin(theta). [Draw the picture. You should see this easily.] Therefore, at B, the tangential component of F is all of
F. Now return to point A, and discard the radial component of F (which makes no torque anyway). The situation is now as follows: A force Fsin(theta) at a radius R produces exactly the same torque as a force F at a radius Rsin(theta).
Therefore, as you go from A to B, the increase in torque that would be caused by changing Fsin(theta) to F must be exactly compensated by a decrease in the torque when R changes to Rsin(theta). In other words, increasing F by some factor is compensated by decreasing R by the same factor. Since it is obvious that torque is proportional to F (pushing harder must obviously accelerate the object more), increasing F by a factor x must increase the torque by a factor x and so decreasing R by the same factor x must decrease the torque by a factor x. This is the definition of being proportional to R.
Hope this helps!