Quantifying Acceleration in Non-Perpendicular Torque Systems

In summary, the conversation discusses the confusion about statical and dynamic behavior and the calculation of acceleration when a body is distant from the rotational axis. The participants also mention the role of a motor and controlling system in maintaining a constant acceleration. The importance of understanding the component of gravitational force acting perpendicular to the radius is emphasized.
  • #1
Andrea Vironda
69
3
Hi,
I'm making some confusion about statical and dynamic behavior.
if i have a body distant from the rotational axis, i have a static torque for maintaining it on position (i suppose ##\vec g \perp \vec r##).
But if a supply more torque i accellerate the body. How can i quantify this accelleration? it will be not constant because ##\vec g## is no more perpendicular to ##\vec r##.
i would obtain ##C(\theta)=mg(\theta)r+I\ddot \theta##. How can a motor mantain a constant accelleration?
 
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  • #2
Andrea Vironda said:
How can a motor mantain a constant accelleration

I don't understand the problem - a suitable choice of motor and control system will do anything that you want it to do .

Is your question really meant to be about how to determine the variable driving torque needed to maintain a constant angular acceleration ?
 
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  • #3
Nidum said:
I don't understand the problem - a suitable choice of motor and control system will do anything that you want it to do .

Is your question really meant to be about how determine the variable driving torque needed to maintain a constant angular acceleration ?
i would to know if my expression in ##C(\theta)## is correct
 
  • #4
What do you intend your expression C(theta) to represent?

Where does a motor enter into the problem as originally described?
 
  • #5
hi, ##C## is a torque, ##coppia## in italian language. It's the torque expression as ##f(\theta)##
 
  • #6
You need to understand that g is not a function of theta; g is a constant. What you need is the component of g that acts perpendicular to the radius. Draw a diagram with labels and it should all become clear.
 

Related to Quantifying Acceleration in Non-Perpendicular Torque Systems

1. What is torque on a horizontal axis?

Torque on a horizontal axis refers to the rotational force applied to an object around a horizontal axis. It is the product of the force applied and the distance from the axis of rotation.

2. How is torque on a horizontal axis calculated?

To calculate torque on a horizontal axis, you need to multiply the force applied by the distance from the axis of rotation. The formula for torque is: torque = force x distance.

3. What is the unit of measurement for torque on a horizontal axis?

The unit of measurement for torque on a horizontal axis is Newton-meters (Nm) in the metric system and foot-pounds (ft-lb) in the imperial system.

4. How does the direction of the force affect torque on a horizontal axis?

The direction of the force applied affects the direction of torque. When the force is applied perpendicular to the axis of rotation, the torque will be maximum. But when the force is applied in the same direction as the axis of rotation, the torque will be zero.

5. What are some real-life examples of torque on a horizontal axis?

Some real-life examples of torque on a horizontal axis include opening a door, using a wrench to loosen a bolt, and kicking a soccer ball. In all these cases, the force is applied perpendicular to the axis of rotation to create torque.

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