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Rotational Kinetic Energy

Rotational Kinetic Energy

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Rotational kinetic energy is simply the energy a rotating object has. It's a type of kinetic energy that comes from an object's rotation kinetic object has related to how it rotates. So, whenever an object spins, it has rotational kinetic energy. This energy is important to understand because it helps us understand how objects move and interact with each other. If you're interested in physics or engineering, you'll definitely want to learn more about rotational kinetic energy!

Rotational Kinetic Energy Formula

Translational kinetic energy, also known as Et, is calculated using a simple formula. The formula uses an object's mass (m) and its translational velocity (v). On the other hand, rotational kinetic energy uses a similar formula, but with a different velocity component. The main difference between the two types of kinetic energy formulas lies in the way they measure velocity. So, if you're studying physics or want to learn more about energy, make sure to take a closer look at the formulas for translational and rotational kinetic energy!

A merry-go-round and planets in the solar system are examples of objects with rotational kinetic energy.
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).

Translational velocity in rotational motion
Translational velocity in rotational motion

When it comes to rotating motion, new variable is needed to fully describe it. This variable is called angular velocity, and it takes into account both the object's translational velocity and its radius. The formula for angular velocity is shown below. It's important to note that angular velocity can also be expressed in terms of period T in seconds or frequency f in Hertz. This is particularly helpful when dealing with periodic motion So, if you're interested in physics, make sure to pay attention the concept of angular velocity!

 

Angular velocity in rotational motion
Angular velocity in rotational motion

In order to calculate rotational kinetic energy (Er), we need to use the formula for translational kinetic energy (Et) and substitute in the appropriate variables. This means that we replace the translational velocity (v) with the angular velocity (ω) multiplied by the radius (r). The full formula for Er is shown below:

Er = 1/2 * m * (ω * r)^2

Here, m is the mass of the object, ω is the angular velocity, and r is the radius of the object. By using this formula, we can calculate the amount of rotational kinetic energy an object has.

It's important to note that the relation between translational and angular velocity can be expressed as v = ω * r. If we substitute this relation into the formula for Er, we get:

Er = 1/2 * m * v^2

This formula looks familiar, as it is the same as the formula for translational kinetic energy. However, it is important to remember that the two formulas are different because they use different velocity components. For rotational motion, we need to use the angular velocity and radius, while for translational motion, we use the straightforward velocity.

By expanding the brackets in the formula for Er, we get:

Er = 1/2 * m * ω^2 * r^2

This expanded formula shows the relationship between the mass, angular velocity, and radius in determining the rotational kinetic energy of an object. 

Moment of inertia and rotational kinetic energy

When dealing with a fixed rotating body, we can simplify our calculations by using the moment of inertia (I in place of the mass. The moment of inertia is a measure of a body's resistance to rotational movement, and it depends on both the mass and how that mass is distributed around the axis of rotation. Specifically, the moment of inertia can be expressed as the product of the mass (m) and the perpendicular distance (r) from the axis of rotation. The formula is shown below:

I = m * r^2

By substituting I for m * r^2 in the formula for rotational kinetic energy, we can simplify the equation and see that it has the same mathematical form as the formula for translational kinetic energy:

Er = 1/2 * I * ω^2

This formula is particularly useful for situations where we have a fixed rotating body with a known moment of inertia. By using this equation, we can easily calculate the amount of rotational kinetic energy possessed by the object. It's important to note that the moment of inertia will vary depending on the shape and distribution of the object's mass. So, make sure to take into account these factors when calculating the moment of inertia. 

Ratio of rotational to translational kinetic energy

(a) To find the rotational kinetic energy of the fan blades, we first need to convert the rotation rate from rpm to rad/s. We have:

70 rpm = 70/60 rev/s = 7/6 rev/s = 7/6 * 2π rad/s ≈ 7.33 rad/s

Next, we can calculate the moment of inertia of each blade using the formula for a thin rod:

I = (1/12) * m * l^2

Here, m is the mass of the blade and l is its length. Since each blade weighs 1 kg and is 0.5 m long, we have:

I = (1/12) * 1 kg * (0.5 m)^2 = 0.0104 kg m^2

To find the total moment of inertia of all three blades, we multiply by the number of blades:

I_total = 3 * 0.0104 kg m^2 = 0.0312 kg m^2

Finally, we can use the formula for rotational kinetic energy to find the energy of the rotating blades:

Er = 1/2 * I_total * ω^2 = 1/2 * 0.0312 kg m^2 * (7.33 rad/s)^2 = 0.93

So, the rotational kinetic energy of the blades 93 Joules.

(b) To find the translational kinetic energy of the fan, we can use the formula for translational kinetic energy:

Et = 1/2 * m * v^2

Here, m is the total mass of the fan and blades, which is 10 kg + 3 kg = 13 kg. The translational velocity is given as 0.5 m/s. So, we have:

Et = 1/2 * 13 kg * (0.5 m/s)^2 = 1.625 J

To find the ratio of translational to rotational kinetic energy, we divide the translational energy by the rotational energy:

Et/Er = 1.625 J / 0.93 J ≈ 1.75

So, the ratio of translational to rotational kinetic energy is approximately 1.75. This indicates that most of the kinetic energy of the is used to rotate its blades, rather than to move it linearly.

Rotational Kinetic Energy Examples

To find the moment of inertia of the disk, we can use the formula for the moment of inertia of a solid disk:

I = (1/2) * m * r^2

Here, m is the mass of the disk and r is its radius. Substituting the given values, we get:

I = (1/2) * 2 kg * (0.5 m)^2 = 0.5 kg m^2

To find the rotational kinetic energy, we can use the formula derived earlier:

Er = 1/2 * I * ω^2

Here, I is the moment of inertia of the disk that we just calculated, and ω is the angular velocity that we found earlier. Substituting the values, we get:

Er = 1/2 * 0.5 kg m^2 * (36 rad/s)^2 = 324 J

So, the rotational kinetic energy of the disk is 324 Joules.

For the second question, to find the total energy of the ball, we need to find both its translational and rotational kinetic energies and add them together.

First, we can use the formula for the moment of inertia of a sphere to find the moment of inertia of the ball:

I = (2/5) * m * r^2

Here, m is the mass of the ball and r is its radius, which is given as 0.4 m. Substituting the given values, we get:

I = (2/5) * 0.3 kg * (0.4 m)^2 = 0.0192 kg m^2

Next, we can find the translational kinetic energy of the ball using the formula:

Et = 1/2 * m * v^2

Here, m is the mass of the ball, which is given as 0.3 kg, and v is its horizontal velocity, which is given as 10 m/s. Substituting the given values, we get:

Et = 1/2 * 0.3 kg * (10 m/s)^2 = 15 J

Finally, we can find the rotational kinetic energy of the ball using the formula:

Er = 1/2 * I * ω^2

Here, I is the moment of inertia of the ball that we just calculated, and ω is the angular velocity that is given as 5 rad/s. Substituting the given values, we get:

Er = 1/2 * 0.0192 kg m^2 * (5 rad/s)^2 = 0.24 J

The total energy of the ball is the sum of its translational and rotational kinetic energies:

E_total = Et + Er = 15 J +0.24 J 15.24

So, total energy the ball when it leaves the hand is 15.24 Joules.

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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