A grindstone of mass 75.0 kilograms consists of a solid disk of radius 0.280 meters. A force of magnitude 180 newtons is applied tangentially to the disk and negligible friction resists the motion of the grindstone. What is the magnitude of the torque exerted by the applied force? What is the magnitude of the angular acceleration of the grindstone due to the applied force? A frictional force of magnitude 20.0 newtons opposes the applied force on the grindstone at a distance of 1.50 centimeters from its axis. What is the magnitude of the grindstone’s angular acceleration taking this friction into account?
This is a three-part problem where we’re given some vital information about our scenario. Let’s record both what we’re given and what we’re asked for in each of the three parts.
We’re told that we have a solid grindstone and that the mass of that grindstone — we’ll call it 𝑚 — is 75.0 kilograms. We’re also told the radius of this grindstone; that value — we’ll call 𝑟 — is 0.280 meters. A force is applied tangentially to the grindstone where the magnitude of that force is 180 newtons. We’ll call that force 𝐹 sub 𝑎 for applied force.
That’s the information we’ve been given. And in part one, we’re asked to solve for the torque that is exerted on the grindstone. We’ll represent that by the lowercase Greek letter 𝜏.
In part two, we’re asked to solve for the magnitude of the angular acceleration that the grindstone experiences under this applied force. We’ll call that 𝛼.
And in part three, a frictional force is introduced that opposes the applied force on our grindstone. We’re told that force is exerted at distance of 1.50 centimeters from the axis of rotation of the grindstone. We’ll call that distance 𝑟 sub 𝑓 for the radius of the frictional force. And we’re told the magnitude of that frictional force, which we’ll call 𝐹 sub 𝑓, is 20.0 newtons.
Under the condition of this frictional force being included, in part three, we’re asked to solve for the new angular acceleration of the grindstone. We’ll call that 𝛼 sub 𝑓 to represent the fact that friction is now involved.
We begin with the first part, solving for torque. Let’s recall the relationship between force, radius, and torque. Torque is defined as 𝑟, the distance from the axis of rotation to where the force is applied, cross 𝐹, the force involved.
In our case, 𝑟 and 𝐹 are applied perpendicularly to one another. That means that as we solve for the magnitude of the torque, we only need to multiply 𝐹 sub 𝑎 times 𝑟.
When we plug in these values, 180 newtons for 𝐹 sub 𝑎 and 0.280 meters for 𝑟, and then multiply these terms together, we find a value for torque of 50.4 newton meters.
That’s how much torque 𝐹 sub 𝑎 applies to the grindstone.
Now let’s move on to part two of our problem, solving for 𝛼, the angular acceleration of the grindstone due to the applied force. As we do this, let’s recall Newton’s second law and in particular the rotational version of this law.
The relationship for Newton’s second law that we’re familiar with is that force is equal to mass times acceleration. The rotational version of the second law or the rotational analogue states that torque is equal to 𝐼, the moment of inertia, times 𝛼, the angular acceleration. The form of the equation is the same, but we’ve now replaced linear terms with rotational terms.
We’re going to use this rotational version of the second law to solve for the angular acceleration of our grindstone 𝛼. Torque is equal to the moment of inertia times 𝛼.
If we divide both sides of our equation by the moment of inertia 𝐼, that term cancels out on the right side of our equation, leaving us with a result 𝛼 is equal to the torque divided by the moment of inertia.
In part one, we solved for the torque 𝜏. For the moment of inertia, that depends on the shape of our object as well as the axis about which it rotates. In our case, we have a disk rotating about an axis that goes through the center of the grindstone.
The moment of inertia 𝐼 for a shape like this that rotates like this is equal to one-half the mass of the disk multiplied by the radius of the disk squared. So we replace 𝐼 in our equation with one-half 𝑚𝑟 squared.
Since we’ve solved for 𝜏 already and we were given 𝑚 and 𝑟 as part of the problem statement, let’s plug in for those values now. 𝜏 is 50.4 newton meters, 𝑚 is 75.0 kilograms, and 𝑟 is 0.280 meters.
When we enter these numbers in our calculator, we find a value for 𝛼, the angular acceleration of the grindstone, of 17.1 radians per second squared. That’s the angular acceleration of the grindstone under the applied force 𝐹 sub 𝑎.
We’re now ready to move on to part three of our problem. In this part, we’ve introduced our frictional force 𝐹 sub 𝑓, which opposes the motion caused by the applied force 𝐹 sub 𝑎.
We want to recalculate 𝛼, the angular acceleration of our grindstone, under the influences of these two forces. The equation that we’ll use to do this remains torque equals 𝐼𝛼. And just like as with part two, we can rearrange this equation so that it reads 𝛼 equals torque divided by 𝐼.
In this case, we’re solving for 𝛼 sub 𝑓, the angular acceleration of the grindstone when the frictional force is present. That means we’ll no longer use 𝜏 that we solved for in part one. Instead, we’ll use a 𝜏 or torque that we can call 𝜏 sub 𝑓, which represents the new net torque on the grindstone under the influence of the frictional force.
As we look at this equation, 𝐼, the moment of inertia of the grindstone, hasn’t changed; that’s still one-half 𝑚𝑟 squared. But 𝜏 has changed. What is 𝜏 sub 𝑓? 𝜏 sub 𝑓 is the net torque that acts on the grindstone under the influence of 𝐹 sub 𝑎 and 𝐹 sub 𝑓.
As we calculate that, let’s define a rotational direction which we’ll call positive. Let’s say that motion that rotates from left to right or counterclockwise as we look down the top of the grindstone is in the direction we’ll call positive.
We saw earlier that the applied force 𝐹 sub 𝑎 acts perpendicularly to 𝑟. Likewise, the frictional force 𝐹 sub 𝑓 acts perpendicularly to 𝑟 sub 𝑓.
This means that as we calculate 𝜏 sub 𝑓, we need only multiply these pairs of values. 𝜏 sub 𝑓 is equal to 𝑟 times 𝐹 sub 𝑎 minus 𝑟 sub 𝑓 times 𝐹 sub 𝑓. This is the net torque that acts on the grindstone under the influence of 𝐹 sub 𝑎 and 𝐹 sub 𝑓.
We’ve been given all the values for these terms. We can now plug them in one by one. 𝑟 is given as 0.280 meters, 𝐹 sub 𝑎 is 180 newtons, 𝑟 sub 𝑓 is given as 1.50 centimeters, which is equal to 1.50 times 10 to the negative two meters, and 𝐹 sub 𝑓 is 20.0 newtons.
So we now have an expression for 𝜏 sub 𝑓 that includes both forces acting on the grindstone that give it torque.
When we enter these numbers into our calculator, we get a value of 50.1 newton meters. Notice that this value differs from 𝜏. It’s less because we now have a frictional force opposing the rotation of our grindstone.
Now let’s look back at our equation for 𝛼 sub 𝑓, the angular acceleration of the grindstone under the influence of both the frictional and the applied force. We now know the value of 𝜏 sub 𝑓 and we were given 𝑚 and 𝑟, so let’s plug those values into that equation to solve for 𝛼 sub 𝑓.
We see that 𝛼 sub 𝑓 equals 50.1 newton meters divided by one half times 75.0 kilograms multiplied by 0.280 meters squared. When we enter these numbers into our calculator, we find a value for 𝛼 sub 𝑓 of 17.0 radians per second squared.
That’s the angular acceleration of our grindstone under the influence of 𝐹 sub 𝑎 and 𝐹 sub 𝑓.