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Rotational Kinetic Energy Formula
The formula of translational kinetic energy (Et) is as follows, where m is mass and v is translational velocity.
\[E_t = \frac{1}{2} \cdot m[kg] \cdot v^2 [m/s]^2\]
While the formula of rotational kinetic energy is very similar to the formula of translational kinetic energy, they differ with respect to the velocity component of the equation.
Figure 1. A merry-go-round and planets in the solar system are examples of objects with rotational kinetic energy.
When we are studying the rotational motion of objects, we can observe that the linear velocity is different for every single point on a rotating cycle of a body about its axis. The reason for this is that linear velocity is a vector quantity, which, in rotational motion, is always tangential to the circle of the motion. Hence, it is always changing direction. This is shown in figure 2, where the velocity of a body varies (v1, v2) at two different time periods (t1, t2).
Figure 2. Translational velocity in rotational motion. Source: Oğulcan Tezcan, StudySmarter.
Therefore, a new variable, called angular velocity, is needed to describe rotating motion more precisely. This variable is related to the magnitude of the translational velocity v and the radius r, as shown in the equation below. It is also useful to note that the angular velocity can also be expressed in terms of period T in seconds or frequency f in Hertz. The latter relation is especially useful for periodic motion.
\[v = \omega \cdot r \quad \omega = \frac{2 \pi}{T} = 2 \pi ƒ\]
Figure 3. Angular velocity in rotational motion. Source: Oğulcan Tezcan, StudySmarter.
To obtain the rotational kinetic energy (Er), we need to substitute angular velocity into the kinetic energy formula (Et), where m is the mass, ω is the angular velocity, r is the radius, and v is the translational velocity.
\[E_t = \frac{1}{2} \cdot m \cdot v^2\]
The relation between translational and angular velocity can be expressed as:
\[v=\omega \cdot r\]
If we substitute translational velocity with the given relation, we get:
\[E_r = \frac{1}{2} \cdot m \cdot (\omega r)^2\]
Expanding the brackets, we get the following for Er:
\[E_r = \frac{1}{2} \cdot m [kg] \cdot \omega^2 [rad/s]^2 \cdot r^2 [m]^2\]
Moment of inertia and rotational kinetic energy
In the case of a fixed rotating body, where we can assume that the mass is concentrated in a single point rotating about a fixed axis, we can use the moment of inertia as an equivalent to its mass.
The moment of inertia (I) is a body’s resistance to rotational movement, which can be expressed as the product of its mass m, and the perpendicular distance r from the axis of rotation, as shown below.
\[I = m[kg] \cdot r^2[m]^2\]
We can further simplify the formula of rotational kinetic energy derived above by substituting the mass and radius with the moment of inertia. It can be seen from the equation below that linear and rotational kinetic energy formulas have the same mathematical form.
\[E_r [J] = \frac{1}{2} \cdot m[kg] \cdot r^2[m]^2 \cdot \omega^2 [rad/s]^2 = \frac{1}{2} \cdot I \cdot \omega^2\]
Ratio of rotational to translational kinetic energy
The ratio of rotational to translational kinetic energy is the rotational kinetic energy over the translational kinetic energy, as shown below, where Et is the translational kinetic energy while Er is the rotational energy. The total kinetic energy in a system that is moving both linearly and rotationally is the sum of linear kinetic and rotational energy.
\[E_{total} = E_r + E_t\]
This ratio is used in cases where an object is rolling or moving linearly with translational kinetic energy and also rotationally with rotational kinetic energy. In order to find the fraction of kinetic energy of an object that is rotational, we have to divide the rotational kinetic energy over the total kinetic energy. To find the fraction of kinetic energy that is translational, we divide the translational energy over the total kinetic energy.
\[E_r = \frac{E_r}{E_r + E_t}; \space E_t = \frac{E_t}{E_r + E_t}\]
A fan weighing 10kg has three blades, where each blade is 0.5 m long and weighs 1kg. The blades are rotating about an axis that is perpendicular to their length. The moment of inertia of each blade can be found using the formula of a thin rod, where m is the mass and l is the length of each rod.
\[I_{blade} = \frac{m_{blade} \cdot r^2}{3}\]
a) What is the rotational kinetic energy of the blades when they are rotating at a rate of 70rpm?
b) What is the translational kinetic energy of the fan when it moves at 0.5 m/s horizontally? Find the ratio of translational to rotational kinetic energy.
Solution (a)
We utilise the rotational kinetic energy formula derived above.
\[E_r = \frac{1}{2} \cdot I \cdot \omega^2\]
However, the rotation rate was given in rpm instead of rad/s, as required in the formula. Therefore, the rotational speed must be converted into rad/s. One rotation per minute is equal to 2π radians per 60 seconds.
\[\omega = \frac{70 rpm}{1 min} \cdot \frac{2 \pi rad}{1 rev} \cdot \frac{1 min}{60 s} = 7.33 rad/s\]
Then, we can calculate the moment of inertia of each blade using the formula provided.
\[I_{blade} = \frac{m \cdot r^2}{3} = \frac{1 kg \cdot (0.5 m)^2}{3} = 0.0833 kgm^2\]
We multiply by the number of blades to find the moment of inertia of all blades.
\[I = 3 \cdot 0.0833 kgm^2 = 0.25 kgm^2\]
Lastly, we substitute the value found into the expression for rotational kinetic energy.
\[E_r = \frac{1}{2} \cdot I \cdot \omega^2 = \frac{1}{2} \cdot 0.25 kgm^2 \cdot (7.33 s^{-1})^2 = 6.72 J\]
Solution (b)
We substitute the given values into the equation for translational kinetic energy.
\[E_t = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} \cdot 10 kg \cdot (0.5 m/s)^2 = 1.25 J\]
To find the ratio of translational to rotational energy, we divide the translational energy by the rotational energy.
\[\frac{E_t}{E_r} = \frac{1.25 J}{6.72J} = 0.186\]
This ratio indicates that most of the kinetic energy of the fan is used to rotate its blades.
Rotational Kinetic Energy Examples
A disk with a radius of 0.5 m and a mass of 2 kg is rotating with a translational speed of 18 m/s. Find the moment of inertia and the rotational kinetic energy.
We begin by using the relation concerning translational and linear velocities in order to find angular velocity.
\[v = \omega \cdot r\]
If we substitute the given variables in the equation above, we get the following value for angular velocity:
\[\omega = \frac{v}{r} = \frac{18 m/s}{0.5 m} = 36 rad/s\]
In order to calculate the rotational kinetic energy, we first calculate the moment of inertia of the disk:
\[I = mr^2 = 2 kg \cdot (0.5 m)^2 = 0.5 kgm^2\]
By substituting the moment of inertia in the rotational kinetic energy formula, we get:
\[E_r = \frac{1}{2} \cdot I \cdot \omega^2 = \frac{1}{2} \cdot 0.5 kgm^2 \cdot (36 rad/s)^2 = 324 J\]
A 0.3 kg ball is thrown into the air with a horizontal velocity of 10.0 m/s. It is rotating at a rate of 5 rad/s. The formula of the moment of inertia of the ball is given by the formula below, where m is the mass, and r is the radius of the ball that is equal to 0.4 m.
\[I_{ball} = \frac{2}{5} \cdot m \cdot r^2\]
What is the total energy of the ball when it leaves the hand?
We utilise the formula of the moment of inertia.
\[I_{ball} = \frac{2}{5} \cdot m \cdot r^2 = \frac{2}{5} \cdot 0.3 kg \cdot (0.4 m)^2 = 0.0192 kgm^2\]
The rotational kinetic energy is found by substituting the moment of inertia into the formula.
\[E_r = \frac{1}{2} \cdot I \cdot \omega^2 = \frac{1}{2} \cdot 0.0192 kgm^2 \cdot (5 rad/s)^2 = 0.24 J\]
The translational kinetic energy is found by substituting the given values of mass and translational velocity in the translational energy formula.
\[E_t = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} \cdot 0.3 kg \cdot (10 m/s)^2 = 15J\]
The total energy is found by the sum of rotational and translational energy.
\[E_{total} = E_r + E_t = 0.24 J + 15 J = 15.24 J\]
Rotational Kinetic Energy - Key takeaways
Rotational kinetic energy is the energy of a rotating body.
The rotational kinetic energy equation has the same form as the linear kinetic energy equation.
Rotational kinetic energy can also be expressed in terms of the moment of inertia of a body.
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Frequently Asked Questions about Rotational Kinetic Energy
What is the rotational kinetic energy of the earth, which has a radius of 6371 km and a mass of 5.972 ⋅ 1024 kg?
The earth completes one rotation about its axis in 24 hours. Converting the period into seconds 86400 sec and using the formulas ω= 2 / T, I= 2/5 m⋅r2 and Er=0.5⋅I⋅ω^2, the rotational kinetic energy of the earth can be calculated as 2.138⋅1029 J.
What is the equation for rotational kinetic energy?
The equation used to calculate rotational kinetic energy is Er=0.5⋅I⋅ω2, where Er is the rotational kinetic energy, I is the moment of inertia, and ω is angular velocity.
How to find rotational kinetic energy without a radius?
Using the moment of inertia, if it has been provided, we can determine this by applying the rotational kinetic energy formula or using the translational to rotational kinetic energy ratio Et /Er.
What fraction of kinetic energy is rotational?
We can find the ratio of translational to rotational energy by dividing Et/Er.
What is the rotational kinetic energy definition?
Rotational kinetic energy is the kinetic energy of a rotating body.
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