In translational dynamics, torque or moment is a vector quantity required to rotate an object about a turning point. Mathematically, it is the product of a force acting on a point and the perpendicular distance from the force to the turning point. The torque is measured in Newton-metres, as shown in the equation below, where F is the force and d the perpendicular distance.
\[T = F[N] \cdot d[m]\]
The principle of moments in translational dynamics states that when an object is in equilibrium, the sum of clockwise moments about a turning point is equal to the sum of anticlockwise moments about a turning point. So where does angular acceleration come into place? Let's find out!
Torque and angular acceleration: Meaning
The torque can also be expressed in terms of rotational dynamics, where it is needed to produce an angular acceleration on an object causing the object to rotate. Translational torque is the product of perpendicular distance and the tangential force acting on a point. The torque of the rotating object shown in figure 1 can be found using Newton’s second law and linear torque.
Figure 1. Torque and angular acceleration in rotational dynamics. Source: Georgia Panagi, StudySmarter.
We express torque T in terms of the translational force Ft in Newtons and the radius r in metres. We then apply Newton’s second law. In the object shown above, we get an expression for translational force in terms of mass and acceleration, where m is the mass and a the acceleration in m/s2.
\[T = F_t \cdot r \cdot \sin{\theta}\]
Where \(\sin{\theta} = 1 \text{ as } \theta = 90^\circ \space F = ma \Rightarrow F_t = ma_t\)
We utilise the translational acceleration at in relation to the derived equation, which expresses the translational acceleration in terms of the radius, and we substitute it into the equation of the translational force.
\[a_t = a \cdot r \qquad F_t = m \cdot (a \cdot r)\]
We then use the torque formula noted above but substitute the derived translational force expression. For an angle of 90 degrees, sinθ is equal to one. The torque is, therefore, equal to the product of mass, acceleration, and squared radius, as shown below:
\[T = (mar) \cdot r \cdot \sin{\theta} = m \cdot r^2 \cdot a\]
The expression that was derived can be written in terms of the moment of inertia I measured in kgm2 multiplied by acceleration measured in m/s2, as the moment of inertia is the mass of an object multiplied by the squared distance to the axis of rotation.
\[T = (m \cdot r^2) \cdot a \qquad T = I \cdot a\]
What is the torque formula for rotational dynamics?
As it was derived above, the torque for rotational motion is defined as the product of the moment of inertia and angular acceleration, as shown below, where T is the torque measured in Newton-metres. I is the moment of inertia, which is the rotating object’s tendency to resist angular acceleration. Angular acceleration α is the rate of change of angular velocity.
\[T[Nm] = I[kg \space m^2] \cdot a [rad/s^2]\]
where \(I[kg \space m^2] = m \cdot r^2\)
It can be derived from the formula that the torque depends on the moment of inertia and hence on the distribution of the mass of the object and its distance to the centre of rotation. The angular acceleration is proportional to the magnitude of torque.
The moment of inertia is the rotational equivalent of force about an axis in translational dynamics, and the angular displacement is the equivalent of the linear displacement in rotational dynamics.
Torque is also equal to the rate of change of angular momentum if the mass is conserved. This is derived below, where ω is the angular velocity in rad/s and L the angular momentum. We use the previous formula of torque in terms of the moment of inertia and substitute the acceleration with the rate of change of angular velocity. That gives us the rate of change of angular momentum, which is equal to the torque, as shown below.
\[T [Nm] = I \cdot a \qquad T = I \cdot \frac{d\omega}{dt} \qquad L = I \cdot \omega \qquad T = \frac{dL}{dt}\]
If two or more forces act on an object, the total torque (\(\tau\)) is the vector sum of the torques, as shown below, where n number of torques are present.
\[\tau = \sum T = T_1 + T_2 + ... + T_n\]
Direction of torque
As torque is a vector quantity, both magnitude and direction are needed to define it. The direction of torque can be found using the right-hand rule, where four fingers of the right hand are pointed in the direction of the force F that is applied. The direction of the torque is the same as the direction of the thumb.
Figure 2. Direction of torque when the force is upwards. Source: Georgia Panagi, StudySmarter.
An example is shown in figure 2, where the force applied is upwards, and the resulting torque is shown below. If the force applied is downwards, the resulting torque is in the opposite direction, as shown in figure 3.
Figure 3. The direction of torque when the force is downwards. Source: Georgia Panagi, StudySmarter.
If the direction of the torque with respect to the point of rotation is clockwise, the torque can be assumed to be negative and vice versa because the angles in an anticlockwise direction are considered positive.
A wheel rotates about axis A. A tangential force of 30 N is applied at the edge of the wheel, which has a radius of 40 cm. The wheel accelerates linearly from rest, reaching a speed of rotation of 10 rad/sec in 8 sec. Determine the moment of inertia of the wheel.
We begin using the torque equation.
\(T = F \cdot r = 30 N \cdot 0.4 m = 12 Nm \qquad T = I \cdot a \Rightarrow I = \frac{12}{a}\)
To determine the angular acceleration in rad/s2, we need to find the rate of change in angular velocity.
\(a = \frac{d \omega}{dt} = \frac{10 - 0}{8s - 0s} = 1.25 rad/s^2\)
We then substitute it into the previous equation to find the moment of inertia:
\(I = \frac{12 Nm}{1.25 rad/s^2} = 9.6 kg \space m^2\)
A wheel is shown below in figure 4. Determine the net torque on the wheel about its centre and the angular acceleration of the wheel if the moment of inertia is 20 kgm2. R1 is 5cm, R2 is 12cm, F1 is 15 N (yellow in the figure below), and F2 is 18 N (blue in the figure).
We create a diagram to visualise the problem.
Figure 4. Example of net torque. Source: Georgia Panagi, StudySmarter.
The total torque is the vector sum of the individual torques acting on the object. We analyse each torque individually.
\(T_1 = F_1 \cdot d \cdot \sin \theta = 15 N \cdot 0 m \cdot \sin(20) = 0 Nm \qquad T_2 = F_2 \cdot R_2 = 18 N \cdot 0.12 m = 2.16 Nm \text{ (clockwise)}\)
We can see that the total torque created by the forces is equal to the torque created by F1 since the torque created by F2 is zero. We continue by using the torque and angular acceleration relation. Note that the result will be negative as the moment is clockwise.
\(T = I \cdot a \Rightarrow a = \frac{T}{I} = \frac{2.16 Nm}{20 kg \space m^2} = -0.108 rad/s^2\)
Torque and Angular Acceleration - Key takeaways
Torque for rotational dynamics is the product of the moment of inertia and angular acceleration.
Moment of inertia is the body’s tendency to resist angular rotation.
The magnitude of the moment of inertia is calculated by the sum of the products of the masses that make up a body with the square of their distance from the axis of rotation.
Angular displacement is the equivalent of linear displacement in rotational dynamics.
The magnitude of the net torque is found by the algebraic sum of the torques acting on the object.
Explanations 
Exams 
Magazine 
