Torque

Torque can be of two types – static and dynamic

A static torque does not produce angular acceleration. For example, pushing a closed door is a static torque. You are applying a force on the door, yet it does not rotate about its hinges (axis) because it is closed. Another example is pedaling a cycle at a constant speed.

A dynamic torque produces angular acceleration. For example, the drive shaft of a racing car produces angular acceleration allowing the race car to drive on its track. A wrench is activated by force applied at one end, creating an unbalanced torque on the tightened element.

Torque is the product of force and distance of the point of application of force and the axis of rotation.

Because it is a product of some distance and a physical entity like force, torque is also called moment of force. r is called the length of the moment arm or lever arm. Also, since the rotating object describes a circle, r is the radius. Notice that the longer the lever arm, the higher the torque.

As mentioned before, torque is a vector quantity. Therefore, in vectorial notation, torque is given by

Using the right-hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of \( \vec{r} \), and curl them to the direction of \( \vec{F} \), then the thumb points in the direction of the torque vector \( \vec{\tau} \).

Suppose the angle between \( \vec{r} \) and \( \vec{F} \) is θ, then the above expression expands to

Symbol & Unit

The Greek symbol τ represents torque. The unit of force is Newton, and that of distance is meter. Therefore, the unit of torque is Newtonˑmeter (Nm). The dimension of torque is ML²T^-2.

From the above equation, when \( \theta = 90^o, \tau =rF \). Therefore, torque is highest when the applied force is perpendicular to the lever arm.

The static torque can be easily measured by using the formula. The dynamic torque is challenging to measure. One can measure the bending force and torque using strain gauges at the pivot point.

Newton’s second law for rotational motion states that every object will move with a constant angular velocity unless a torque acts on it. According to the second law for rotation,

Where,

I: Moment of inertia

α: Angular acceleration

Therefore, the torque is directly proportional to the moment of inertia and angular acceleration. The above equation is an essential equation for studying rotational kinematics. Rearranging it, we find that

The greater the torque, the higher the angular acceleration. Also, for a massive object, the moment of inertia is high. Hence, the angular acceleration is low.

Rotational Equilibrium

An object can be subjected to several torques. If the net torque on the object is zero, it is in rotational equilibrium.

Suppose forces F₁, F₂, …, and F_n act at distances r₁, r₂, …, and r_n from the pivot point or fulcrum. Then the torque balance equation is

Power

When an object rotates, there will be energy associated with it. We calculate the power of the object as follows.

Where ω is the angular velocity. The power is the product of the torque and the angular velocity.

Torque is applied to any object where there is a pivot point. Here are some applications.

• Wrench
• Seesaw
• Pendulum
• Steering wheel

Problem 1. A 4 m beam of negligible weight balances in equilibrium with a fulcrum placed 1.5 m from its left end. If a force of 60 N is applied on its right end, how much force must be applied to the left?

Solution

Given F₁ = ?, r₁ = 1.5 m, F₂ = 60 N, r₂ = 4 m – 1.5 m = 2.5 m

Torque on the left lever arm (τ) = r₁ x F₁ = 1.5 m x F₁

Torque on the right lever arm (τ) = r₂ x F₂ = 2.5 m x 60 N = 150 Nˑm

Since the beam is balanced, the two torques cancel out. Therefore,

τ₁ = τ₂

=> 1.5 m x F₁ = 150 Nˑm

=> F₁ = 150 Nˑm /1.5 m = 100 N

Problem 2. One side of a seesaw carries a 32 kg mass four meters from the fulcrum and a 36.5 kg mass two meters from the fulcrum. To balance the seesaw, what mass should be placed eight meters from the fulcrum on the side opposite the first two masses?

Solution

Given m₁ = 32 kg, r₁ = 4 m, m₂ = 36.5 kg, r₂ = 2 m, m₃ = ?, r₃ = 8 kg

Using the torque balance equation

r₁ˑF₁ + r₂ˑF₂ + r₃ˑF₃ = 0

=> 4 m ˑ 32 kg ˑ g + 2 m ˑ 36.5 kg ˑ g – 8 m ˑ m₃ ˑ g = 0

m₃ = (4 m ˑ 32 kg + 2 m ˑ 36.5 kg)/8 m = 25.1 kg