Is it easier to move or rotate an object? [closed]

This is a problem I'm facing on designing a moving bookcase for my home, I just don't have enough physics background to tackle the problem.

Which of the following requires more force/effort?

• A bookcase with a pivot point in one of the back corners that is rotated 180° to its new position (therefore we now face the back of the bookcase).

• A bookcase that is pushed/pulled to the same relative position but with a diagonal push (guided by one rail at the top).

In both cases the bookcase will be supported by equivalent fixed caster wheels aligned with the direction of movement or rotation, with the pivot and the rail acting only as guide.

I don't think it should matter, but the bookcase will be about 1 ft deep and 3 ft wide, I don't have any good source, but I think it would be around 200 kg if fully loaded.

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closed as too localized by Qmechanic♦, Sklivvz♦Jan 3 '13 at 13:28

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If you want to do this rigorously you need to phrase the question more carefully. Assuming we can neglect friction it's possible to move the bookcase with an arbitrarily small amount of work because you can apply a small force for a short time to accelerate the bookcase then wait for it to move into place. So if you're prepared to wait a long enough time you can move the bookcase with as small a force as you want.

I'd guess you're really asking something like: if I want to move the bookcase in a set time $t$ by first pushing to accelerate it then pushing to slow it down again, is the force bigger if I slide it or if I rotate it?

Start by sliding the bookcase. You need to apply a force F until the bookcase has moved a distance $w/2$ then apply the same force in the other direction to slow it again, and you want to do this in some set time $\tau$. The equation you need is:

$$s = \frac{1}{2}at^2$$

$a$ is the acceleration, which is $F/m$, $s$ is the distance moved and $t$ is the time taken. If you just consider the first half of the move, i.e. moving the bookcase a distance $w/2$ in a time $\tau/2$ then we can put these values in the above equation to get:

$$\frac{w}{2} = \frac{1}{2} \frac{F}{m} \frac{\tau^2}{4}$$

and a quick rearrangement tells us that the force $F$ is given by:

$$F_{slide} = 4 \frac{mw}{\tau^2}$$

Working out the force involved in rotation is pretty similar, except that the equation we need is now:

$$\theta = \frac{1}{2}\frac{T}{I}t^2$$

where $\theta$ is the angle moved, $T$ is the torque and $I$ is the moment of inertia. As above we take the first half of the rotation i.e. rotate the bookcase an angle $\pi/2$ in a time $\tau/2$. You also need to know that torque is force times distance from the hinge i.e. $T = Fw$, and the moment of inertia of slab of width $w$ hinged about the edge is $I = mw^2/3$. So substitute all of this in our equation and we get:

$$\frac{\pi}{2} = \frac{1}{2} \frac{Fw}{mw^2/3} \frac{\tau^2}{4}$$

and rearranging this gives:

$$F_{rot} = \frac{4\pi}{3}\frac{mw}{\tau^2}$$

Lets take the ratio $F_{rot}/F_{slide}$:

$$F_{rot}/F_{slide} = \frac{\pi}{3}$$

So the answer is that the force needed to rotate the bookcase is higher than the force needed to slide it by a factor of $\pi/3$.

BUT

All this is dependant on a specific model of how you move the bookcase, I've ignored friction, and in any case the factor of $\pi/3$ is pretty close to unity. I think all we can really say is that it probably doesn't make much difference whether you slide or rotate the bookcase.

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Really nice, thank you :) I still haven't digest it all but that is the answer that I needed. – Luiz Borges Sep 1 '12 at 11:41

Ofcourse rotating is easier but you neet to push longer because rather than pushing r-meters, you actually follow a $pi*r$ (180 degrees) road.

For rotating, the static-friction to count is easier for the points closer to the pivot. For the straight-pushing, each point has equally magnitude of static-friction.

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I thought about it and it looks like it makes sense, but couldn't actually form a rationalle on how that would work (or on how to calculate it). What I thought, was that I could calculate the individual path lenghts of each wheel on the rotation case and somehow compare it to the straight push case, but I don't quite remeber my high school physics and how to work that out. Do you care to elaborate on how I would go on to calculate it (just for fun)? BTW, why would the STATIC friction be smaller (shouldn't it be the kinetic friction)? – Luiz Borges Aug 31 '12 at 23:34
Ok, we can try taking an area*sin(angle) integral in a chat. I dont know how to open a new chat window – huseyin tugrul buyukisik Aug 31 '12 at 23:40
Ok, I just understood why the friction is static when dealing with wheels, but how do I go from the force required to count the static friction and put it into motion to the $F=ma$? What is the force in that case? – Luiz Borges Sep 1 '12 at 0:29
I cant find the exact force withot measuring geometric and material properties of the case. – huseyin tugrul buyukisik Sep 1 '12 at 8:55