# What is the explanation of greater torque having greater "rotatory effect" on a stationary body? [closed]

Why does moving a constant force further from center of mass (and thus increasing torque) increase angular acceleration? I know that the explanation for my question is that during the same angular displacement the same amount of force would be exerted over longer (linear) displacement, hence doing more work, and since conservation of energy is a thing the angular acceleration would increase, but it (at least for me) doesn't explain why moving a weight on a balanced beam makes it move (because when it is not yet moving, no work is being done and I can't mathematically prove that it would move).

I used to ask myself a similar question: How does a load at one end of a lever know how far away along the lever another force is being exerted? Forgive the anthropomorphic phrasing, but I hope you get my drift.

I solved my problem by realising that the lever had to have some internal structure. I considered the case of a lever of length 6$$a$$, pivoted at a point one third of the way along. I envisaged a lever in the form of a pin-jointed lattice of thin weightless struts and ties. The bottom of the lever is a thin horizontal rod of length 6$$a$$, resting on the pivot. A distance $$a$$ above the bottom rod is another thin horizontal rod. Between the two rods are 6 rods of length $$\sqrt 2 a$$, angled at 45°, each at 90° to its neighbour(s), to form a lattice running from one end of the bottom rod to the other. Two of the angled rods are one side of the pivot; the other four are on the other side; the top horizontal rod needs to be only $$4a$$ long.

You will find simply by applying force resolution at each joint that if a force $$W$$ is applied downwards at the end of the long arm of the lever, a force $$2W$$ is needed at the end of the short arm, in order to have equilibrium.

I then convinced myself that this result is independent of the particular internal structure chosen. This is almost certainly a very eccentric method of establishing the law of the lever (principle of moments) but I found it instructive and convincing! • Thank you for your answer. Since the most force resolution I've ever done is decomposing a force into x, y and z components, I can't wrap my head around it. This is how I got stuck: if I decompose the force to each beam, then the net force on each beam should be the initial force divided by number of beams on each side and the net force on each side should be net force on each beam times number of beams of that side. This is when I get that both forces would be the same. Nov 19, 2022 at 20:12
• @Henry05 I'll append details in an hour or two – technology permitting. Nov 19, 2022 at 21:10
• Clever derivation! With a little calculus, this can actually demonstrate that you can derive all of the laws of rotational motion from linear motion, and likewise all of the laws of linear motion from rotational motion. Typically we break motion apart into linear and angular motion simply because the math is easier if we do so, not because there's some fundamental difference between them. Nov 20, 2022 at 16:51

You can experience this effect if you go to your local playground and use the seesaw. Ask a friend to sit at a fixed position, then sit down at different leverages. The seesaw will rotate.

This effect is not "due to conservation of energy". However, we can use "energy considerations" to calculate the point of equilibrium. This is motivated in the following picture: The left picture sketches the initial state and the right picture the equilibrium state.

• Thank you for your answer. I apologize, I have not been clear when I asked for proof. I am convinced that it would happen. What I don't understand is why. The only explanation I could find involved work and it was not satisfactory for me, since it can't prove why it starts to move (as far as I can tell). Nov 19, 2022 at 16:44
• @Henry05 Why don't you tell us where you found the explanation being work. Cite a reference Nov 19, 2022 at 16:58
• @BobD It was explained like this at my elementary and high school. Of course, there's also the answer (which is more common) that simply just reiterates the formula - which doesn't explain anything, only describes it. qr.ae/pvdevq physics.stackexchange.com/questions/242713/…. (which is literally the answer google gives you when you ask this question) Nov 19, 2022 at 17:16
• @Henry05 I thought you were talking about work done moving the weight horizontally when you stated " moving a weight on a balanced beam ", not the work done due to the net torque resulting from moving the weight Nov 19, 2022 at 17:56
• This seems like another way to visualize the center if gravity explanation from one of the other answers, and I like it. At the same time, it immediately leads to a question about why the center of gravity is important. Nov 20, 2022 at 6:26

When the beam is stationary, the center of mass due to the weights on both sides is directly over the fulcrum and there is zero net torque as measured about the fulcrum. If one weight is moved, the c.m. shifts so it is no longer over the fulcrum. There is then a net torque about the fulcrum due to the weights, considered to be acting at the center of mass. The more the weight is moved, the more the c.m. is shifted away from the fulcrum therefore it has more effect so that's how the radius comes into play.

We can imagine a limit situation, where a mass $$m$$ is very far from the pivot point, that is when the distance $$r \to \infty$$. And the rod joining it to that point has a very small mass compared to $$m$$.

At a first approximation, the movement is linear for an observer close to $$m$$, and by the second Newton's Law: $$\mathbf F = m\mathbf a$$.

If we suppose the force orthogonal to the radius $$r$$: $$Fr = mar = mr^2\frac{a}{r} \implies \tau = I\alpha$$

• Thank you for your answer. If I understand correctly, you used Fr, or the definition of torque (or rather a special case), in the equation, but that is what I struggle with - why is that equation a thing; meaning - why do we have to multiply force by radius to find it's "rotational accelerative effect"? Knowing this would also explain for me why there's an r^2 in the moment of inertia. Nov 19, 2022 at 20:21
• It is a matter of names. It is possible to say that $Fr$ is proportional to $\frac{a}{r}$, and that the constant of proportionality is $mr^2$. The names: torque, angular acceleration and moment of inertia are only convenient conventions. Nov 19, 2022 at 22:09

Does it though? Increasing the arm (for a constant force) always increases force, but the angular acceleration doesn't necessarily have to. Consider two point masses connected by a rod of length $$2r$$ (see figure below). To each point mass a force of strength $$F$$ is applied always at 90 degrees to the rod. I choose this setup to simplify the situation as much as possible.

In this case the angular acceleration can be found simply by \begin{align} \alpha&=\frac{\tau}{I}\\ &=\frac{2rF}{2mr^2}\\ &=\frac{F}{mr} \end{align} We find that, for constant force, the angular acceleration actually decreases! This makes sense if you imagine a very long and imagine how fast it rotates. • Thank you for your answer. I have already thought of this example (or similar), and it was one of the main reasons why this problem confused me, because it proved the opposite (which happens way too often to me with physics). Nov 20, 2022 at 18:14

Maybe look at it the other way around -- in terms of work done.

The force involved comes from gravity.

When the two weights are balanced... if {the weight that we are going to move outwards (away from the fulcrum)} were to go down... the work done by gravity would be the same on the two weights, except that one would be negative.

If we move the weight away from the fulcrum -- say to 2 * {the original distance} -- then... if we imagine that it is moved down 1 unit of distance... the other weight is moved up 1/2 a unit of distance.

p.s. It feels tangentially relevant to say that... when doing calculations for, e.g. pushing down on a table, one imagines that the table is pushing back with equal force. This is strictly a fudge, to get the correct value for the equation, that results in the table not moving (up nor down).

Its already implied if we accept that a balanced beam wont rotate. Lets look at the case of an essentialy massless beam, and place the two equal weights $$w$$ a distance $$d$$ to the right and to the left of the fulcrum, respectively. Of course the beam wont rotate, otherwise it would mean that nature disobeys an almost trivial symmetry. If we then move the weights so they are both directly above the fulcrum, it still wont rotate, because of the same trivial symmetry. We deduce that a force acting directly through the fulcrum, wont cause the beam to rotate. Lets go back to the balanced beam with the weights a distance $$d$$ from the fulcrum. But now look at the situation with the left weight as the reference point. Seen from there, the normal force from the fulcrum is acting at a distance $$d$$ and with a force $$2w$$ upwards, and the rightmost weight is acting at a distance $$2d$$ with a force $$w$$ downwards. Now remove the fulcrum, but replace it with an equal force of magnitude $$2w$$ acting upwards, i.e. the same situation. Put the fulcrum directly under the leftmost weight, i.e. at the current reference point. The leftmost weight cant contribute, since its acting through the fulcrum (if you want, you can elongate the beam leftwards so that the beam is symmetric about the new fulcrumpoint, but that doesnt change anything as we consider the beam essentially massless). It follows that for nature to obey the trivial symmetry, it has to be the case that a force $$2f$$ at a distance $$r$$ from a point, must have the same tendency to rotate a beam as a force $$f$$ at a distance $$2r$$ from the point.