The magnitude of torque on a particle of mass 1 kg is 2.5 N-m about the origin. If the force acting

Q: Why do we use the cross product for torque?

Torque is a rotational force, and its magnitude depends on the perpendicular component of the force relative to the position vector. The cross product naturally extracts this perpendicular component using the sine of the angle between them.

Q: Does the mass of the particle affect the torque?

No, the mass of the particle is extra information in this specific problem. Torque only depends on the force applied, the distance from the pivot, and the angle between them.

Q: What if the angle was 0 or 180 degrees?

If the angle was 0 or 180 degrees, the sine of the angle would be zero, resulting in zero torque. This means the force is acting directly along the line of the position vector, causing no rotation.

Q: How do we convert degrees to radians?

To convert degrees to radians, multiply the angle in degrees by $\pi / 180$. For example, $30^\circ \times (\pi / 180) = \pi / 6$ radians.

Q: Can torque be negative?

The magnitude of torque is always positive. However, the direction of torque (clockwise or counterclockwise) can be represented by a positive or negative sign depending on the chosen coordinate system convention.

Question

The magnitude of torque on a particle of mass $1	ext{ kg}$ is $2.5	ext{ N-m}$ about the origin. If the force acting on it is $1	ext{ N}$ and the distance of the particle from the origin is $5	ext{ m}$, then the angle between the force and the position vector is (in radian)

Options:

(A) $\frac{\pi}{8}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

Sigma Percentile · Accepted Answer

The concept of torque is fundamental in understanding rotational motion. Just as force causes linear acceleration, torque causes angular acceleration. In this problem, we are given the magnitude of torque, the force applied, and the distance from the origin, and we need to find the angle between the force and the position vector.

### Visualizing the Setup

Imagine a particle of mass $m = 1 	ext{ kg}$ located at a distance $r = 5 	ext{ m}$ from the origin. A force $F = 1 	ext{ N}$ acts on this particle. The force vector and the position vector are not necessarily aligned; they have an angle $	heta$ between them. This angle is what determines how effectively the force can cause rotation around the origin.

### The Master Equation

The magnitude of torque $	au$ is defined as the magnitude of the cross product of the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$. Mathematically, this is expressed as:
$$|	au| = |\mathbf{r} 	imes \mathbf{F}| = rF \sin	heta$$

This equation tells us that torque depends on three factors:
1. The distance from the pivot point ($r$).
2. The magnitude of the applied force ($F$).
3. The angle between the force and the position vector ($	heta$).

### Substituting the Values

We are given the following values:
- $|	au| = 2.5 	ext{ N-m}$
- $F = 1 	ext{ N}$
- $r = 5 	ext{ m}$

Substituting these values into our master equation, we get:
$$2.5 = 5 	imes 1 	imes \sin	heta$$

### Solving for the Angle

Now, we simply need to solve for $\sin	heta$:
$$2.5 = 5 \sin	heta$$
$$\sin	heta = \frac{2.5}{5} = \frac{1}{2}$$

We know from basic trigonometry that the angle whose sine is $1/2$ is $30^\circ$. In radians, this is:
$$	heta = \frac{\pi}{6} 	ext{ rad}$$

**Conclusion:** The angle between the force and the position vector is $\frac{\pi}{6}$ radians. Notice how the mass of the particle ($1 	ext{ kg}$) was not needed to solve the problem. This is a common trick in physics problems to test if you truly understand which variables are relevant to the concept being tested!

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Visualized Solution

$Visualizing the Setup$

$Formula for Torque$

$Substituting the Values$

$Solving for sin θ$

$Finding the Angle θ$

The Sigma Insight: Torque and Angular Momentum

Visualizing the Setup

The Master Equation

Substituting the Values

Solving for the Angle

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for sinθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Angle θ