# Video: Finding Changes in Rotational Kinetic Energy of a Rotating Object

A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. What is the rotational kinetic energy of the grindstone when it is rotating at 1.5 × 10³ rpm? After the grindstone’s motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work-energy theorem to determine how many turns the grindstone makes before it stops.

09:24

### Video Transcript

A uniform cylindrical grindstone has a mass of 10 kilograms and a radius of 12 centimetres. What is the rotational kinetic energy of the grindstone when it is rotating at 1.5 times 10 to the third rpm? After the grindstone’s motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 5.0 newtons. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops.

Let’s begin by highlighting some of the important information given. We’re told that the grindstone has a mass of 10 kilograms and a radius of 12 centimetres and that at the outset, the grindstone is rotating at 1.5 times 10 to the third revolutions per minute. After the motor of the grindstone is turned off, a knife blade is pressed perpendicularly against the grindstone’s edge with a force of 5.0 newtons. And we’re told the coefficient of kinetic friction between the grindstone and the blade is 0.80.

In part one of this problem, we want to solve for the rotational kinetic energy of the grindstone. In part two, we want to solve for what we’ll call capital 𝑇, the total number of rotations the grindstone goes through as it comes to a stop. Let’s draw a diagram of this scenario.

The grindstone rotates about an axis through its centre. It has a mass, we’ll call 𝑚, of ten kilograms and a radius, we’ll call 𝑟, which is 12 centimetres. We’re told it spins at a rate of 1.5 times 10 to the third revolutions per minute.

To solve for the rotational kinetic energy of the grindstone, let’s recall the relationship for rotational kinetic energy. The rotational kinetic energy of an object KE sub 𝑟 is equal to one-half the object’s moment of inertia 𝐼 multiplied by its angular speed 𝜔 squared. In the case of the cylinder rotating about its central axis, the cylinder’s moment of inertia is one-half its mass times its radius squared.

We can use this relationship for rotational kinetic energy for our particular scenario. The rotational KE of the grindstone is one-half times one-half 𝑚 𝑟 squared times 𝜔 squared. In the problem statement, we’re given the mass and radius of the grindstone. But recall that 𝜔 is an angular speed in units of radians per second. Since we’ve been given an angular rotation rate in revolutions per minute, let’s convert that now into units of radians per second.

In one minute, there are 60 seconds and there are two 𝜋 radians in one full revolution. When we cancel out common units, we see we’re left with radians per second, which are the units of 𝜔.

So let’s multiply these three fractions together. 𝜔 equals 157.08 radians per second. Now, we know 𝑚, 𝑟, and 𝜔, all the elements to solve for the rotational kinetic energy of the grindstone.

Let’s insert those values into that equation now. The rotational kinetic energy equals one-quarter times the mass 10 kilograms times the radius in units of metres 0.12 metres squared multiplied by 𝜔 squared. To two significant figures, the rotational kinetic energy of the grindstone is equal to 890 joules. This is its kinetic energy rotating at 1.5 times 10 to the three rpm.

Now that we’ve solved for KE sub 𝑟, let’s look into- now that we’ve found the rotational kinetic energy, let’s look into part two: solving for the number of turns the grindstone would make as it comes to a stop. In part two, what we’re told that a knife is pressed against the edge of the rotating grindstone with the force, we’ll call 𝐹 sub 𝑁, which has a magnitude of 5.0 newtons.

This push creates a frictional force opposing the rotation of the grindstone; we’ll call that force 𝐹 sub 𝑓. Under the influence of this frictional force, the grindstone comes to a stop. And we want to know how many times it turns around before it does. The instructions are to use the work energy theorem to solve this problem. So let’s recall that theorem now.

The work energy theorem tells us that the work done on an object is equal to the object’s change in kinetic energy. Since we’re talking about work, let’s recall another relationship for that term. We’ve seen that work is equal to the force on an object times the displacement of that object; that is for linear work done And for rotational work 𝑊 sub 𝑟, rotational work is equal to 𝜏 torque times 𝜃 angular displacement.

If we combine the work energy theorem with the definition for rotational work, in this scenario the change in kinetic energy of the grindstone is equal to the torque applied to it multiplied by its angular displacement 𝜃. Since the grindstone goes from its rotational kinetic energy we solved for in the first part to a rotational kinetic energy of zero once it comes to a stop, we’ve already solved for ΔKE; that’s equal to KE sub 𝑟.

Looking at the right side of the equation. Let’s recall that torque is equal to the position vector, starting at the axis of rotation called 𝑟, times the tangential force being applied 𝐹. Looking at our diagram, a vector from the axis of rotation to the point where the force is applied is perpendicular to the direction of that force.

So in our situation, torque is equal to the frictional force times 𝑟, the radius of the grindstone. But what is the frictional Force 𝐹 sub 𝑓? Frictional force is given as the coefficient of friction between two surfaces 𝜇 times the normal force 𝐹 sub 𝑁.

And in our problem statement, we’re told that the coefficient of kinetic friction 𝜇 sub 𝑘, in our case, is equal to 0.80. So in our equation for torque, we can substitute in for 𝐹 sub 𝑓. It’s equal to 𝜇 sub 𝑘 times 𝐹 sub 𝑁, the normal force of the knife against the grindstone.

All three of these values are known, so we can insert those values for these symbols and solve for the torque 𝜏: 0.80 times 5.0 newtons times 0.12 metres. This multiplies out to a torque value of 0.48 newton metres.

Now that we know 𝜏 and KE sub 𝑟 from part one, we can return our attention to our equation involving 𝜃, the angular displacement. To isolate 𝜃, let’s divide both sides by 𝜏. 𝜃 is equal to the rotational kinetic energy divided by 𝜏 or 890 joules divided by 0.48 newton metres.

Before we calculate this fraction, let’s recall that 𝜃 is a value in radians. But the answer we seek, 𝑇, is the number of complete revolutions that the grindstone will make. To convert from radians to revolutions, we want to multiply 𝜃 by one divided by two 𝜋 radians. When we convert from radians to revolutions, like we have here, we change from 𝜃 and angular displacement to 𝑇, which represents the number of revolutions the grindstone makes before it comes to a stop.

Entering these values on the right side into our calculator, we find a value for 𝑇 to two significant figures of 290 revolutions. That’s how many times the grindstone will turn before coming to a complete stop.