# B How much force (N) would it take to rotate this object?

1. Oct 16, 2016

### EnjoysLearning

As depicted in the picture.

Sorry, I have minimal experience with physics at all lol, but my paint game is strong right?

It is a 1 kg object with the center of mass perfectly in the center, rotated 90 degrees in 0.1 seconds. What about in 1 second?

http://imgur.com/OIOdZ6f
http://imgur.com/OIOdZ6f

How much force (in Newtons) would it take for 0.1 seconds or 1 second? Would it make a difference if the object is falling/suspended? With/without gravity? Is there any difference Position 1->2 vs Position 2->1?

Relevant equations would be great too, and they'd probably answer multiple (if not all) of my questions at once.

Last edited: Oct 16, 2016
2. Oct 16, 2016

### haruspex

Is this homework?

3. Oct 16, 2016

### EnjoysLearning

No, although I can see how it would seem that way. I've actually only ever taken one physics course, but this is a combination of curiosity as well as something that would help me for my own independent things.

4. Oct 16, 2016

### EnjoysLearning

You could move it there, but I wouldnt have the slightest idea on how to start.

5. Oct 16, 2016

### Staff: Mentor

We need a bit more information in order to help get you started.

Is the object moving in an up-down direction (so gravity forces will come into play)? Or moving on a horizontal surface? Is friction involved? Is the object loose, or constrained by a pivot joint? Does it hit a stop at the end of its movement, or do you need to accelerate and then decelerate it? Are the dimensions arbitrary now, or do you have the actual dimensions and mass of the bar?

EDIT -- Ah, I see it is rotating around it's center of mass?

Last edited: Oct 16, 2016
6. Oct 16, 2016

### Svein

Is the mass uniformly distributed? We need to know in order to calculate the moment of inertia.

7. Oct 16, 2016

### EnjoysLearning

Yes

8. Oct 16, 2016

### EnjoysLearning

Hmm, yes up->down. So gravity will be in play. What amount of force would it take to accelerate to that position in say 0.1 seconds (then how much force to decelerate it to stop it from over rotating).

Arbitrary at the moment. Does it matter if it is uniformly distributed mass?

9. Oct 16, 2016

### Svein

Yes. If almost all mass is in the center, it is much easier to rotate than if almost all mass is in the outer shell.

10. Oct 16, 2016

### Staff: Mentor

If it's rotating about its center of gravity, then gravitational forces cancel out.

Is there much friction at the bearing point in the middle of the bar? Or should we just assume it's frictionless? How will you apply the rotational torque (what you are calling Force) to the bar? Via an axle? Or an external force near one or both ends?

11. Oct 16, 2016

### haruspex

To be clear, is it pivoted at its mass centre or just floating in space?

12. Oct 16, 2016

### EnjoysLearning

Floating in space, but the center of mass is in the middle since the mass is uniformly distributed. Assume frictionless at the center and that the force is being applied externally at one end, as depicted in my magnificent paint picture

13. Oct 16, 2016

### haruspex

Ok.
We need to determine the moment of inertia of the block about its mass centre. This represents its resistance to being rotated faster, in the same way that the mass of an object is its resistance to being accelerated.
For a uniform block mass M, length L, width W, being rotated about the third dimension, it is M(L2+W2)/12.
The force applied exerts a torque about the mass centre. The torque is the force multiplied by the distance from its line of action to the mass centre. In this case that is FL/2. At least, that is what it is initially. One more question needs to be answered: as the block rotates, does the force keep pointing in the same direction, or does it turn around with the block so that it is always at right angles to the block?

The resulting angular acceleration is the torque divided by the moment of inertia. Assuming the force stays at right angles to the block, that gives constant angular acceleration $\alpha=\frac{6FL}{M(L^2+W^2)}$. Finally, we can apply the SUVAT formulae for constant acceleration. The angle through which it will turn in time t (measured in radians) is $\frac 12\alpha t^2$. To turn through $\frac 12\pi$, the time is $\sqrt{\frac{\pi}A}$, or about $1.77/\sqrt A$.

On the other hand, if the force keeps pointing in a constant direction, the torque diminishes. Suppose the block has rotated through an angle $\theta$ so far. The torque is now $FL\cos(\theta)/2$, so $\alpha=\frac{6FL\cos(\theta)}{M(L^2+W^2)}$. Alpha is the second derivative of theta wrt time, so $\ddot \theta=\frac{6FL\cos(\theta)}{M(L^2+W^2)}$. Writing A for the constant $\frac{6FL}{M(L^2+W^2)}$, this gives the differential equation $\ddot \theta=A\cos(\theta)$. Multiplying both sides by $\dot\theta$ allows us to integrate this to get $\dot\theta=\sqrt{2A\sin(\theta)}$. (The initial condition that the block is not rotating means there is no constant of integration.)
This leads to an "elliptic integral". Fortunately, you are asking about the time taken to rotate through $\pi/2$, so we have a definite integral with convenient limits. The solution is $K(\frac 12)/\sqrt A$, where K is a "complete elliptic integral" (see http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html). $K(\frac 12)$ is about 1.85, so slightly longer.

Last edited: Oct 16, 2016