A system consists of an annular cylinder of mass 1.0 kilograms with radius 20 centimeters and outer radius 30 centimeters, mounted on a disk of mass 2.0 kilograms and radius 50 centimeters. The system rotates through the centers of the annual cylinder and of the disk at 10 revolutions per second. What is the moment of inertia of the system? What is the system’s rotational kinetic energy?
Let’s start by highlighting some of the vital information given to us in the statement. The system consists of two masses: the first mass is an annual cylinder of 1.0 kilograms with inner radius 20 centimeters and outer radius 30 centimeters; the second mass making up the system is a disk of 2.0 kilograms which itself has a radius of 50 centimeters.
We’re told that the axis of rotation runs through the centers of the annular cylinder and of the disk and that the system overall rotates at 10 revolutions per second. We want to solve for the moment of inertia of the system, we’ll call that 𝐼.
And we also want to find out the rotational kinetic energy of the system, we’ll call that KE sub 𝑟. As a first step in our solution, let’s draw a diagram of this system. Here, we have a diagram of our system. The cylinder shaded in blue sits on top of the disk shaded in gold.
The radius of the solid disk is 𝑟 sub three. The outer radius of the cylinder is 𝑟 sub two, and the cylinder’s inner radius is 𝑟 sub one. Those values are given to us in the problem statement: 𝑟 sub one is 20 centimeters; 𝑟 sub two is 30 centimeters; and 𝑟 sub three is 50 centimeters.
We’re also told the mass of the cylinder and the mass of the disk. The mass of the cylinder we’ve called 𝑚 sub 𝑐, that’s 1.0 kilograms. And the mass of the disk, 𝑚 sub 𝑑, is 2.0 kilograms. In part one of the problem, we’re asked to solve for the moment of inertia of this system.
Since the system consists of two distinct shapes, the cylinder and the desk. We’ll need to calculate the moment of inertia for each of those two shapes and then add them together to find the total moment of inertia, 𝐼. That means we can write 𝐼 as the sum of 𝐼 sub 𝑑, for the moment of inertia of the disk, and 𝐼 sub 𝑐, the cylinder’s moment of inertia.
To solve for the moments of inertia of the disk and the cylinder, we look these shapes up in a table of moment of inertias There we find that for a disk of mass 𝑚 and radius 𝑟, the moment of inertia of the disk is one-half its mass times radius squared.
Likewise when we look up the moment of inertia of an annular cylinder we find that value is equal to one-half the annual cylinder’s mass multiplied by the outer radius squared plus the inner radius squared.
We can replace 𝐼 sub 𝑑 and 𝐼 sub 𝑐 in our overall moment of inertia equation with these new terms. In terms of the values given in our problem statement, 𝐼, the moment of inertia, is equal to one-half the mass of the disk times 𝑟 sub three squared plus one-half the mass of the cylinder multiplied by 𝑟 sub two squared minus 𝑟 sub one squared.
Let’s plug in for the masses and the radii of our system. And as we do that, we’ll replace the radii in centimetres with equivalent radii in meters so that our units are consistent. When we put these values into equation we find that 𝐼 is equal to one-half times 2.0 kilograms times 0.50 meters squared plus one-half times 1.0 kilograms times the quantity 0.30 meters squared plus 0.20 meters squared.
When we enter these values in our calculator, we find a value for 𝐼 of 0.32 kilogram-meters squared. Now that we’ve solved for our systems moment of inertia, let’s move on to solving for its rotational kinetic energy.
We’ve seen before the equation for linear or translational kinetic energy. That equation states that the kinetic energy of an object equals one-half its mass times its speed squared.
The rotational version of that equation, KE sub 𝑟, says that rotational kinetic energy is equal to one-half 𝐼𝜔 squared. So in place of 𝑚 in the linear version, we have 𝐼, the moment of inertia. And in place of 𝑣 in the linear version, we have 𝜔, the rotational speed.
Let’s use this relationship for rotational kinetic energy in our scenario. As we look at this equation, we see that we’ve already solved for 𝐼 in the previous part of the problem. Now we want to solve for 𝜔, the rotational speed of our system.
In the problem statement, we’re told that our system rotates at a rate of 10 revolutions per second. Now at this point it’s important to remember the units of 𝜔, the rotational speed; those units or radians per second.
And we recall that two 𝜋 radians is equal to one revolution. So therefore, we can’t call the 10 revolutions per second value 𝜔 for our system. We’ll need to do a little bit of conversion first.
Since two 𝜋 radians equals one revolution, then that means that 10 revolutions is equal to 10 times two 𝜋 radians.
If we divide both sides of the equation by seconds, we then see that the right side of the equation matches our initial condition of 10 revolutions per second. This means that 10 revolutions per second is equal to ten times two 𝜋 radians per second, or 20𝜋 radians per second.
That number, the number in units of radians per second, is what we’ll call 𝜔, the angular speed, in units of radians per second. We are now ready to plug into equation for rotational kinetic energy. We’ve solved for 𝐼, and we’ve just solved for 𝜔, the angular speed.
When we enter those values in we find that the rotational kinetic energy of the system is one-half 0.32 kilogram-meters squared multiplied by 20𝜋 radians per second squared.
When we enter these numbers into our calculator, we see that the rotational kinetic energy is equal to 620 joules. We use two significant figures in this answer to match with the precision of the values we were given in the problem statement. This value is the rotational kinetic energy of this system.