# B Calculating Linear and Angular Quantities from Applied Force

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1. Oct 26, 2016

### StarWarsNerd

In the image above, a centroid with radius 1 is depicted. F1 is pointing directly at point A (which is the center of the circle), and F2 is pointing directly at point B. The radius for finding the torque would be the perpendicular between the center of the object and the force vector, so r1 would be 0 and r2 would be 1.

From my understanding, F1 will make the circle move in the direction of the force, but it will not rotate. F2 on the other hand, since the radius is not zero, will make it move in the direction of the force and rotate.

So my question is, given a force vector f and a radius r, how do you calculate what the final meters/second and the radians/second of the object will be?

2. Oct 26, 2016

### Khashishi

Are you familiar with torque? What are the equations for torque and angular acceleration?
Calculate the torque associated with F2.
You need the moment of inertia of the circle. If you know the mass distribution, you can calculate this. Try to do this.

In theory, the linear acceleration doesn't depend on where you apply the force, but in practice, it's hard to apply a constant force on something that is rotating.

3. Oct 26, 2016

### StarWarsNerd

• F = m * a
• T = F * r * sin(θ)
• α = Δω / Δt
• I = ¼ * π * r4
So if we have Fn, a force equal to those in the image, but anywhere along the diameter, and the radius would then be rn we get
T = Fn * rn * sin(π/2) = m * a * rn * 1
This is where I get stuck, I am not sure how to proceed.

So if the circle rotates it will travel as far as one that does not rotate?

4. Oct 26, 2016

### jbriggs444

It depends on the details. What factors are you holding constant for the comparison? What factor are you changing?

The same force applied for the same length of time but changing the initial point of application to one that is farther right?

The same force applied over the same vertical distance, allowing the point of application to rotate right and away, but stopping sooner because the point of application rotates away faster than the object as a whole?

The same force applied while the center of mass of the object moves for the same vertical distance, but changing the initial point of application?

The same force applied for the same length of time but with a sticky little roller so that the point of application stays at the same relative position while the object rotates under it?

5. Oct 26, 2016

### StarWarsNerd

A force with the same magnitude/direction, applied for the same period of time, just the x coordinate is moving left or right.

6. Oct 26, 2016

### jbriggs444

Then a momentum/impulse argument can give you the desired answer. Change in momentum = impulse. Impulse = force multipled by duration. Rotation does not enter in. Linear momentum and angular momentum are conserved separately.

7. Oct 26, 2016

### StarWarsNerd

So to calculate the linear velocity I would just take the F * (1/m) * duration? What about the angular velocity? That formula is really tripping me up.

8. Oct 26, 2016

### Khashishi

Your expression for moment of inertia has the wrong units.
You are missing one other equation:
$\tau = I \alpha$
which means
torque = moment of inertia * angular acceleration.

The angular formulas are a lot like the linear formulas, but you have to replace force with torque, mass with moment of inertia, velocity with angular velocity, acceleration with angular acceleration.

9. Oct 26, 2016

### StarWarsNerd

so I = ¼ * m * r2 and τ / I = α → (F * r) / (¼ * m * r2) = (4 * F) / (m * r)
and for the linear component, we take F = m * a → a = F / m
Is that correct? And this would apply for any force, not just tangential or one pointed at the center of mass?

10. Oct 26, 2016

### jbriggs444

Where are you finding the formula for the angular momentum of a thin disc about an axis through and perpendicular to its center?

As I read the original posting, the axis is perpendicular to the disc, not coinciding with a diameter.

11. Oct 26, 2016

### StarWarsNerd

You are correct, the axis of rotation would be the z-axis, meaning I = ½ * m * r2 and α = (2 * F) / (m * r) ?

12. Oct 26, 2016

### jbriggs444

Heh, I suspected you were looking at that wiki article. Yes, the angular acceleration looks right to me.

13. Oct 26, 2016

### StarWarsNerd

You said earlier that linear momentum and angular momentum are conserved separately, so, the angular acceleration does not change the linear acceleration, or do I need to calculate linear using angular?

14. Oct 26, 2016

### jbriggs444

They are separate.

Changes in angular momentum depend on the sum of the external torques. External forces do not enter in as forces -- though they may contribute a torque due to their offset from the axis.

Changes in linear momentum depend on the sum of the external forces. External torques do not enter in. From a linear momentum point of view, a force is a force, no matter where it is applied and how much torque it may contribute.

[Edited to change from "accelerations" to "changes in momentum"]

15. Oct 27, 2016

### StarWarsNerd

To get the direction of the rotation, do you need to recalculate torque, or can α be use to find if it is spinning clockwise or counterclockwise?

16. Oct 27, 2016

### jbriggs444

When you calculate the angular acceleration [in the two dimensional plane], you can pick a sign convention for $\alpha$. For instance, positive for clockwise and negative for counter-clockwise.

17. Oct 27, 2016

### StarWarsNerd

So α does not intrinsically have a cw/ccw direction

18. Oct 27, 2016

### jbriggs444

As soon as you decide upon a sign convention for it, it does.

Normally, I'd expect to use counter-clockwise = positive to match the way angles increase in the first quadrant from 0 = straight right with positive numbers progressively counter-clockwise from there.