Torque about a point
Torque about a Point
Torque, also known as the moment of force, is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear motion, torque is what causes an object to acquire angular acceleration.
Understanding Torque
Torque ((\vec{\tau})) is a vector quantity that measures the tendency of a force to rotate an object about some axis. The point about which the torque is measured can be any point in space, but it is often chosen to be the point of rotation or the center of mass of an object.
The magnitude of torque depends on three factors:
- The magnitude of the force applied ((F))
- The distance from the point to the line of action of the force, which is the lever arm ((r))
- The angle ((\theta)) between the force vector and the lever arm
The formula for torque is given by:
[ \tau = r \cdot F \cdot \sin(\theta) ]
where:
- (\tau) is the torque
- (r) is the lever arm (distance from the axis to the point where the force is applied)
- (F) is the magnitude of the force
- (\theta) is the angle between the force vector and the lever arm
In vector form, torque can be expressed as the cross product of the position vector ((\vec{r})) and the force vector ((\vec{F})):
[ \vec{\tau} = \vec{r} \times \vec{F} ]
Important Points and Differences
| Aspect | Description |
|---|---|
| Direction of Torque | The direction of the torque vector is perpendicular to the plane formed by (\vec{r}) and (\vec{F}). It follows the right-hand rule. |
| Units of Torque | The SI unit of torque is the newton-meter (Nm). |
| Sign of Torque | Torque can be positive or negative depending on the direction of rotation it tends to produce. |
| Equilibrium | An object is in rotational equilibrium when the net torque acting on it is zero. |
| Torque and Angular Acceleration | Torque is directly proportional to angular acceleration, just as force is to linear acceleration. |
Examples
Example 1: Calculating Torque
A force of 10 N is applied at the end of a 0.5 m long wrench, and the force is applied perpendicularly to the wrench. What is the torque about the axis of rotation at the other end of the wrench?
Using the formula:
[ \tau = r \cdot F \cdot \sin(\theta) ]
Since the force is applied perpendicularly, (\theta = 90^\circ) and (\sin(90^\circ) = 1). Therefore:
[ \tau = 0.5 \, \text{m} \cdot 10 \, \text{N} \cdot 1 = 5 \, \text{Nm} ]
Example 2: Torque with an Angle
A force of 20 N is applied at a 30-degree angle to a lever arm that is 0.4 m long. Calculate the torque about the pivot point.
Again, using the torque formula:
[ \tau = r \cdot F \cdot \sin(\theta) ]
[ \tau = 0.4 \, \text{m} \cdot 20 \, \text{N} \cdot \sin(30^\circ) ]
[ \tau = 0.4 \, \text{m} \cdot 20 \, \text{N} \cdot 0.5 ]
[ \tau = 4 \, \text{Nm} ]
Example 3: Vector Cross Product
Consider a force (\vec{F} = (3 \, \text{N}, 4 \, \text{N}, 0)) applied at a point with position vector (\vec{r} = (2 \, \text{m}, 0, 0)). Calculate the torque about the origin.
Using the cross product:
[ \vec{\tau} = \vec{r} \times \vec{F} ]
[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 0 \ 3 & 4 & 0 \ \end{vmatrix} ]
[ \vec{\tau} = (0 - 0) \hat{i} - (0 - 0) \hat{j} + (8 - 0) \hat{k} ]
[ \vec{\tau} = 8 \hat{k} \, \text{Nm} ]
The torque is 8 Nm in the positive (k)-direction, indicating a counterclockwise rotation about the (z)-axis.
In conclusion, torque about a point is a fundamental concept in rotational motion, and understanding how to calculate it is essential for solving problems in physics and engineering. The direction of the torque, its relationship with angular acceleration, and the conditions for equilibrium are crucial for a comprehensive understanding of rotational dynamics.