Work is done in general by a force that is acting on a body, causing a certain displacement of the body. In linear motion, work is equal to the product of the force and the distance. For angular motion, angular work is the work done by an object performing angular motion in order to rotate the object around an axis. Similarly, angular power is the power delivered to a rotating object.
Angular work
Angular work is the equivalent of linear work in angular motion. In linear motion, a force is acting on an object, and work is done on it. Similarly, in angular motion, work has to be done to rotate an object about an axis. Angular force is known as torque and consists of any force which causes an object to turn about some axis. Work is the result of a torque acting over an angle.
Formula derivation for angular work
For linear motion, the work done is equal to the product of the force acting on a body and the linear displacement. For angular motion, we convert the linear displacement into angular displacement, using the arc length and radius relation. Hence, the angular work done is equal to the force ‘F’ multiplied by the angular displacement ‘\(\theta\cdot r\)’, with \(\theta\in [0,2\pi]\) as shown below: \(W_{linear} = Fx; \quad x = \theta \cdot r\)
\(W_{ang} = F(\theta \cdot r)\)
However, for circular motion, it is also known that torque ‘T’ is equal to a force acting on the body multiplied by its radius ‘F•r’. By replacing the torque relation with the angular work that was derived, we conclude that angular work and linear work equations have the same form. Furthermore, the torque is the reciprocal of force in linear work, and angular displacement is the reciprocal of linear displacement.
Angular work, just as linear work, is expressed in the unit of Joules.
\(T = F \cdot r\)
\(W_{ang} [J] = T \cdot \Delta \theta\)
Angular Work and Power: Torque direction
The direction of torque can be found by using the right-hand rule, where we point our fingers towards the direction of rotation. We then extend our thumb upwards. By pointing the fingers in the direction of rotation, the thumb points to a direction perpendicular to the direction of rotation. The direction of the torque is found by the direction of the thumb (see figure 1).
In vertical rotation, the following applies:
- If the direction of rotation is clockwise, the torque is downwards.
- If the direction of rotation is anticlockwise, the torque is upwards.
Figure 1. Direction of rotation and torque. Source: StudySmarter.
For non-vertical rotation, the direction of torque can be found using the right-hand rule. This states that four fingers follow the direction of rotation while the thumb is extended. The direction of the thumb shows the direction of the torque being either upwards or downwards. An example is shown in figure 2 below.
Figure 2. Direction of torque. Source: Georgia Panagi, StudySmarter.
Angular Work and Power: Work-energy theorem
The work-energy theorem states that the total work done from external forces on a body is equal to the change of kinetic energy.
However, the change in kinetic energy must also include the translational kinetic energy ‘TKe’ and the rotational kinetic energy ‘RKe’. In the formula below, W is work, and ΔKe is the change in kinetic energy.
\(W = \Delta Ke = TKe - RKe\)
where \(RKe = \frac{1}{2} \cdot I \cdot \omega^2 [J]\)
Angular power
Power is defined as the rate at which energy is being transferred.
Linear power is the work done over time. Using the previous relations, we know that the work done for linear motion is equal to the product of the force and its linear displacement. Displacement over time equals velocity.
Formula derivation for angular power
For angular motion, we have proved earlier that the torque is the reciprocal of the force in linear motion, while angular velocity is the reciprocal of linear velocity.
Substituting the angular velocity into the equation, we get an equation in terms of torque in which torque, measured in Newtons per metre, is equal to the moment of inertia multiplied by the angular acceleration.
\[P[W] = \frac{w}{t}\]
We then modify the earlier equation for work, which is equal to the product of force and linear displacement. We substitute the force with torque over time, as by definition, the torque is equal to the product of the force and the radius, as seen below.
\[W = F \cdot x \space T = F \cdot r \Rightarrow F = \frac{T}{r} W_{ang} = \Big( \frac{T}{r} \Big) \cdot x\]
Here, x is the linear displacement, which is equal to linear speed multiplied by time. Since we are dealing with angular rotation, the linear speed v has to be replaced with its angular equivalent, which is utilised below where ω is the angular speed.
\[V = r \cdot \omega \cdot t\]
This will be substituted into the linear displacement term, which gives us the expression below.
\[W_{ang} = \Big( \frac{T}{r} \Big) \cdot x \qquad W_{ang} = \Big( \frac{T}{r} \Big) \cdot r \cdot \omega \cdot t\]
We can now substitute this derived work equation for angular rotation into the power equation.
\[P[W] = \frac{W_{ang}}{t} = \frac{\frac{T}{r} \cdot r \cdot \omega \cdot t}{t} = T \cdot \omega\]
At this point, the radius term and time are cancelled out, so we are left with a simpler expression where power is the product of torque T, and angular velocity ω is measured in rad/s2.
A torque of 300 k Nm is applied on a turbine, which is rotating at 15 rad/s. Determine the power needed for the turbine to continue rotating.
Solution:
\(P = T \cdot \omega = 300.000 [k Nm] \cdot 15 [rad/s] = 4.500 MW\)
A rotating wheel with a radius of 0.5m and a mass of 3kg is pulled with a string with a force of 10N for a distance of 0.8 m. Determine the work done.
Solution:
We begin by using the torque equation, as the torque is required to determine the work done.
\(W_{ang}[J] = T \cdot \Delta \theta T = F \cdot r = 10 N \cdot 0.5 m = 5 Nm\)
Then we use the displacement and radius relation to determine the angular displacement in radians, which is required in the work equation. The equation is rearranged to make θ the subject before we substitute the values obtained in the initial work equation.
\(x = \theta r \quad \theta = \frac{x}{r} = \frac{0.8}{0.5} = 1.6 rad \qquad W = T \space \theta = 5 Nm \cdot 1.6 = 8 J\)
Angular Work and Power - Key takeaways
Angular work is the work done by a torque acting on a body multiplied by the angular displacement.
Angular power is the rate of change of energy transfer in angular motion.
Rotational dynamics obey the same rules as linear dynamics, as linear formulas have the same form as rotational ones.
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