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Motion of System of Particles and Rigid Body - Moment of a Force and Torque

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Rigid Body is defined as a system of particles in which the distance between any two constituent particles remains constant, regardless of the external force applied.

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The Moment of a Force (or Torque) is the measure of the tendency of a force to cause rotation of a body about a specific point or axis.

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Torque is a vector quantity. Its direction is determined by the Right-Hand Thumb Rule and is perpendicular to the plane containing the position vector r⃗\vec{r} and the force vector F⃗\vec{F}.

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The magnitude of torque depends on the magnitude of the force FF, the distance rr from the axis of rotation, and the angle θ\theta between them.

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A Couple consists of two equal and opposite parallel forces acting along different lines of action. The total linear force of a couple is zero, but it produces a pure rotational effect.

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The Principle of Moments states that for a body in rotational equilibrium, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about that same point: ∑τclockwise=∑τanticlockwise\sum \tau_{clockwise} = \sum \tau_{anticlockwise}.

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In SI units, the unit of Torque is N⋅mN \cdot m. Its dimensional formula is [ML2T−2][M L^2 T^{-2}].

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Torque is the rotational analogue of force; while force causes linear acceleration, torque causes angular acceleration α\alpha.

📐Formulae

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

τ=rFsin⁡θ\tau = r F \sin \theta

τ=Force×Perpendicular distance (lever arm)\tau = Force \times \text{Perpendicular distance (lever arm)}

τ=Iα\tau = I \alpha

Moment of a Couple=Force×Arm of the Couple\text{Moment of a Couple} = \text{Force} \times \text{Arm of the Couple}

τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}

💡Examples

Problem 1:

A force F⃗=(2i^+3j^−k^) N\vec{F} = (2\hat{i} + 3\hat{j} - \hat{k}) \text{ N} acts at a point whose position vector is r⃗=(i^−j^+2k^) m\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) \text{ m}. Calculate the torque about the origin.

Solution:

The torque is given by the cross product τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. τ⃗=∣i^j^k^1−1223−1∣\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix} τ⃗=i^[(−1)(−1)−(2)(3)]−j^[(1)(−1)−(2)(2)]+k^[(1)(3)−(−1)(2)]\vec{\tau} = \hat{i}[(-1)(-1) - (2)(3)] - \hat{j}[(1)(-1) - (2)(2)] + \hat{k}[(1)(3) - (-1)(2)] τ⃗=i^(1−6)−j^(−1−4)+k^(3+2)\vec{\tau} = \hat{i}(1 - 6) - \hat{j}(-1 - 4) + \hat{k}(3 + 2) τ⃗=−5i^+5j^+5k^ N⋅m\vec{\tau} = -5\hat{i} + 5\hat{j} + 5\hat{k} \text{ N}\cdot\text{m}

Explanation:

We use the vector cross product method to find the torque when the force and position are given in component form. The magnitude of the resulting vector gives the total torque, and its direction is the axis of rotation.

Problem 2:

A nut is opened by a wrench of length 20 cm20 \text{ cm}. If a force of 50 N50 \text{ N} is applied at the end of the wrench at an angle of 90∘90^\circ, calculate the torque produced.

Solution:

Given: F=50 NF = 50 \text{ N}, r=20 cm=0.2 mr = 20 \text{ cm} = 0.2 \text{ m}, and θ=90∘\theta = 90^\circ. Using the formula τ=rFsin⁡θ\tau = r F \sin \theta: τ=0.2×50×sin⁡(90∘)\tau = 0.2 \times 50 \times \sin(90^\circ) Since sin⁡(90∘)=1\sin(90^\circ) = 1: τ=10 N⋅m\tau = 10 \text{ N}\cdot\text{m}

Explanation:

When the force is applied perpendicularly to the lever arm, the torque is maximized because sin⁡(90∘)\sin(90^\circ) is 11. The length must be converted to meters to maintain SI units.

Moment of a Force and Torque - Revision Notes & Key Formulas | ICSE Class 11 Physics