Calculating Angular vs Lateral Acceleration of Objects Suspended in Zero Gravity

  • Thread starter xpoisnp
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In summary: The reason the angular velocity will be greater is as I explained above.In summary, when a force is applied to an object suspended in zero gravity, it can be divided into two components: the force at the center of mass, which causes linear acceleration, and the moment at the center of mass, which causes angular acceleration. The percentage of the force applied to the object's angular acceleration is determined by the magnitude of the moment and its perpendicular distance from the center of mass. This can be calculated using vectors or by replacing the applied force with an equivalent force and moment at the center of mass. When the force is applied off-center, both linear and angular velocities will change, but the work done by the force will remain the same.
  • #1
xpoisnp
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Imagine you had a metal cube suspended in a zero gravity vacuum. If you were to poke the cube precisely in the middle of one of its faces exactly towards its center of mass, the cube would accelerate forward directly away from you with no angular acceleration.

If you were to move your hand towards the edge of the face and poke in the SAME forward direction, now the cube would gain some amount of spin and would therefore not move away from you as fast.

My question is: given a force applied to an object suspended in zero gravity, what portion of that force will be applied to the object's angular acceleration and what proportion to its lateral acceleration?

I'm most interested in two-dimensional examples, so you can imagine the same experiment performed on a square with evenly distributed mass on a 2D plane, where all the force vectors being applied to the square are in that plane. With that, any force vector can be divided into two component vectors: the vector pointing directly towards (or away from) the object's center of mass from the point of contact and the vector perpendicular to it. Obviously the component vector pointing directly towards (or away from) the object's center of mass won't ever contribute to angular acceleration, so my question then becomes what percentage of a vector perpendicular to the vector between the point of contact and the object's center of mass will be applied to its angular acceleration?

I expect the answer is as simple as "100% will be applied to angular acceleration" but I haven't convinced myself of that yet. And if it is that simple, I imagine adding friction makes things more interesting, so given coefficients of kinetic friction for angular and lateral movement, is it possible to calculate the portion of force perpendicular to the center of mass of the object that gets applied to angular acceleration versus lateral acceleration?

Oh and a WONDERFUL THANKS to anyone able to help me with this. I've been mulling this over for a while and just need good convincin' more than anything else!
 
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  • #2
xpoisnp said:
With that, any force vector can be divided into two component vectors: the vector pointing directly towards (or away from) the object's center of mass from the point of contact and the vector perpendicular to it. Obviously the component vector pointing directly towards (or away from) the object's center of mass won't ever contribute to angular acceleration, so my question then becomes what percentage of a vector perpendicular to the vector between the point of contact and the object's center of mass will be applied to its angular acceleration?

Splitting the force into components like that was a good idea, but it doesn't answer your question easily. For example, you could apply a force that has no component towards the center of mass, so "splitting the force" wouldn't lead anywhere.

The way to solve this is replace the applied force by the equivalent system of a force (in the same direction) at the center of mass, plus a moment about the center of mass. In 2D, the moment = (magnitude of the force) times (perpendicular distance to the center of mass). In 3D, the idea is the same but it's probably easier to use vectors - see https://engineering.purdue.edu/~aprakas/CE297/CE297-Ch3.pdf for example

The force at the CM then produces linear acceleration, and the moment at the CM produces angular acceleration.
 
  • #3
xpoisnp said:
Imagine you had a metal cube suspended in a zero gravity vacuum. If you were to poke the cube precisely in the middle of one of its faces exactly towards its center of mass, the cube would accelerate forward directly away from you with no angular acceleration.

If you were to move your hand towards the edge of the face and poke in the SAME forward direction, now the cube would gain some amount of spin and would therefore not move away from you as fast.
No. If you apply the same force over the same time, the linear veloctiy will be the same.
 
  • #4
xpoisnp said:
If you were to move your hand towards the edge of the face and poke in the SAME forward direction, now the cube would gain some amount of spin and would therefore not move away from you as fast.

That is wrong (as the previous replies have said) but it may seem paradoxical that it's wrong, because somehow "the same force" is providing "the same linear acceleration" plus some angular acceleration as well. The translational kinetic energy is always the same, but there is also some rotational kinetic energy, depending where the force is applied.

The explanation is: if the force is applied off-center, the point where it is applied moves further in a given amount of time (because the object rotates as well as translates), so the work done by the force (= force times distance) is greater. That is where the "extra" rotational kinetic energy comes from.

If you push by moving your hand at the same speed but at different positions, you will apply less force if you push "off-center", and so the translational velocity will be less.
 
  • #5


I can provide you with some insights into the calculation of angular and lateral acceleration of objects suspended in zero gravity. First, it is important to understand that in a zero gravity vacuum, there is no external force acting on the object, therefore the only forces acting on the object are the ones applied by the experimenter. This means that the total force applied to the object will be divided between its angular and lateral acceleration.

In order to calculate the angular and lateral acceleration of an object, we need to use the equations of motion. The equation for angular acceleration is given by α = τ/I, where α is the angular acceleration, τ is the torque applied to the object and I is the moment of inertia. On the other hand, the equation for lateral acceleration is given by a = F/m, where a is the lateral acceleration, F is the force applied to the object and m is its mass.

Now, let's consider the two scenarios mentioned in the content. In the first scenario, the force is applied precisely in the middle of one of the faces of the cube, towards its center of mass. In this case, the torque applied to the object will be zero, as the force is applied at the center of mass. Therefore, all of the force will be applied to the object's lateral acceleration and none to its angular acceleration. This explains why the cube accelerates directly away from you without any spin.

In the second scenario, the force is applied towards the edge of the face, but still in the same forward direction. In this case, the torque applied to the object will be non-zero, as the force is not applied at the center of mass. This means that some of the force will be applied to the object's angular acceleration, resulting in the cube gaining some spin and not moving away from you as fast.

To answer your question about the proportion of force applied to angular and lateral acceleration, it is not a fixed percentage and depends on the location and direction of the force applied. As shown above, if the force is applied at the center of mass, then all of it will be applied to lateral acceleration and none to angular acceleration. However, if the force is applied at a distance from the center of mass, then a portion of it will be applied to angular acceleration and the remaining to lateral acceleration.

The presence of friction will also affect the proportion of force applied to angular and lateral acceleration. Friction will create a torque that opposes the applied
 

1. How do you calculate angular acceleration in zero gravity?

To calculate angular acceleration in zero gravity, you need to know the moment of inertia of the object and the net torque acting on it. The formula for angular acceleration is:
α = net torque / moment of inertia

2. How does angular acceleration differ from lateral acceleration in zero gravity?

Angular acceleration is the rate at which an object's angular velocity changes, while lateral acceleration is the rate at which an object's linear velocity changes. In zero gravity, the absence of a gravitational force means that objects do not experience linear acceleration, but can still experience angular acceleration due to rotational forces.

3. Can an object experience both angular and lateral acceleration in zero gravity?

Yes, an object can experience both angular and lateral acceleration in zero gravity. For example, a spinning top will experience both types of acceleration simultaneously as it spins and moves in a straight line.

4. How is the moment of inertia of an object calculated?

The moment of inertia of an object depends on its shape and mass distribution. It is calculated by summing the products of each infinitesimal mass element of the object and its distance from the axis of rotation squared. The exact formula varies depending on the shape of the object.

5. Can objects in zero gravity experience angular acceleration without any external forces?

No, according to Newton's First Law of Motion, an object will maintain its state of rest or uniform motion unless acted upon by an external force. In zero gravity, there is no external force to cause angular acceleration, so the object will continue to spin at a constant angular velocity.

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