If you don’t know what you’re doing, solving rotational motion and torque problems for your physics class can get ugly.
Here’s the scenario:
You’re finally starting to get comfortable with the idea of velocities, acceleration, force, and momentum. You can do this in both X and Y: projectiles, blocks sliding down slopes, ropes and pulleys, etc.
But then, we start rotating, and everything becomes “angular this” and “angular that.”
Now once you have this stuff down, you’ll see how similar rotational motion is to linear motion, but until that point it can be a little tough to wrap your head around the connection.
So to help with that, below I go through a solution to a rotational motion problem pulled from a Physics 1 exam.
Let’s jump in.
Rotational Motion and Torque Problem Statement
A Yo-Yo of mass m has an axle of radius b and a spool of radius R. It’s moment of inertia can be taken to be I=1/2mR^2 and the thickness of the string can be neglected. The Yo-Yo is released from rest.
(a) What is the tension in the cord as the Yo-Yo descends?
(b) Use conservation of energy to find the angular velocity of the Yo-Yo when it reaches the bottom of the string.
(c) What happens to the Yo-Yo at the bottom of the string?
From MIT OCW
What is the tension in the cord as the Yo-Yo descends?
Following our problem solving process, first things first, let’s restate the problem and identify exactly what we’re trying to find.
Here we know the mass, radius of the Yo-Yo axle, and the overall radius of the Yo-Yo; however, they’re represented in symbolic form.
Brief side note: I know a lot of you are more comfortable with numbers, but symbolic problems are actually a blessing in disguise because they force you to really understand and establish the relationships between the physical variables at play instead of immediately jumping into plugging in stuff.
We also know what the moment of inertia (MOI) of the Yo-Yo is, and that it’s initially released from rest. We’re also assuming that the string has no thickness, so that won’t be a complicating aspect of this problem.
Taking a guess to establish a baseline first, I’m estimating that the force due to tension is greater than the force due to gravity. (Note: I realize how ridiculously incorrect this is, but that’s okay at this point. Making terrible conceptual mistakes and then going back to correct them later is what really sharpens your physics intuition.)
Diagram & Variables
Then I’m re-creating the diagram for the problem and labeling all of the known and unknown variables on the Yo-Yo that I think I’m going to need to use in the problem.
Again, we’re assuming that our initial velocity V0 is 0, and gravity g is acting downwards. The force due to tension Ft is acting up the string in the opposite direction of gravity, and this is the unknown variable we’re looking to solve for in Part (a).
Free Body/Torque Diagram & Equations
Next step is to create both a free body diagram (FBD), as well as a torque diagram (TD).
Any time you’re asked to determine force, acceleration, torque, etc. in a rotational motion problem, it’s a good bet that you’re going to need to set up some equations that come from these two types of diagrams. Plus, it’s usually helpful anyway to analyze the problem in this was just to get a sense of what’s going on.
From our FBD, we know that the Yo-Yo is moving downwards with a linear acceleration a, and this comes from the resultant y-direction force that comes out of the opposing forces of the string tension and gravity. Based on this we can re-arrange to get an initial equation (1) that describes our tension force; however, a is not given in the problem so we’ll need some additional equations.
From our torque diagram then, we can establish a torque balance equation that describes the relationship between the tension force Ft and the rotational acceleration alpha. This is found by using the two main torque equations (equals 90-degree force times distance AND equals MOI times rotational acceleration). From this we can establish equation (2) that involves the tension force; however, this equation also has alpha as an unknown (we’re not given angular acceleration of the Yo-Yo).
So finally, to relate these two different equations to each other, we can establish and equation that describes the relationship between linear and angular acceleration – equation (3).
Now that we have 3 equation and 3 unknowns, we simply need to do some substitution and algebra to solve for the tension force Ft in terms of the variables that are given in the problem (m, g, R, b).
So now going back to compare this answer to my guess (which was 2mg) I see how wrong I was with my initial intuition. Our solution is telling us that the force of tension is LESS than the force due to gravity (clearly, if the Yo-Yo is accelerating downwards it has to be), and this force is related to the geometry of Yo-Yo, more specifically the relationship between the overall radius, and the radius of the axle R^2/(R^2 + 2b^2).
Rotational Motion and Torque Part (b)
Use conservation of energy to find the angular velocity of the Yo-Yo when it reaches the bottom of the string.
Alright so in part (b) we’re shifting our focus off of force and torque and onto energy and velocity.
Because the problem statement is giving me a hint that we need to use conservation of energy, I can remember that this means that we’re keeping the same amount of energy in the system, but it’s going to be moving from potential energy to kinetic energy.
Also because I know that the equation for potential energy due to gravity is always mgh and the kinetic energy equation always has to do with a squared velocity, my guess is a square root of some ratio between gravity, height, and the geometry of the Yo-Yo.
Diagram & Variables
For my diagram in part (b) I want to show myself what the motion we’re talking about is going to look like.
We know that the Yo-Yo initially starts at rest from the original problem description (so both V0 and w0 are equal to zero), so that’s our energetic starting point. And we also know that it’s going to drop some height h (related to how long the string is) as it picks up speed, eventually reaching a final linear and angular velocity (Vf and wf) when it hits the end of the string.
Everything else is the same as in part (a)
Energy State Diagram & Equations
Next I’m constructing an Energy State Diagram, which is critical for comparing two different physical states that involve the same amount of energy (since we know this is a conservation of energy situation).
Here I’m depicting both the height and velocities involved in each state, because they will determine what the potential and kinetic energy is, which added together gives us the total energy of that state.
In energy State 1, like we said earlier, both the linear and angular velocity is zero, so we have no kinetic energy. However, the Yo-Yo is released at some height h above the point at which it reaches the end of the string, so this gives us a potential energy of mgh, which in turn gives us our total energy for State 1 – equation (1).
Note: Because we’re not given any information about the length of the string in this problem, we’ll assume that h is a known quantity that we can use in our final answer.
Then for energy State 2, we know that our height h has now reached zero (we’re at the end of the string), and the linear and angular velocity has reached it’s maximum – the values we’re looking for in the problem. Because of this, the total energy for State 2 is all kinetic. And here we remember that kinetic energy can come from both linear motion as well as rotational motion, and to find the total energy you just add those together to get equation (2): (1/2)mVf^2 + (1/2)Iwf^2.
Now we have the total energy of both states, and can use conservation of energy to set those two energies equal to each other for equation (3); however, we still can’t solve for our angular velocity wf yet because we also have vf in the equation which we don’t know.
So to fix this, just like we did in part (a), we can establish a relationship between the linear motion quantity and the rotational (angular) motion quantity, which in this case is velocity. This gives us equation (4), and we now have 4 equations and 4 unknowns.
Here again, we’re just using algebra to isolate wf in terms of variables that are known (g, h, b, R) using the 4 equations that we just established, which gives us our final answer.
Now I was actually pretty close on my guess here. It was a square root with gh on the top, and some geometric relationship on the bottom; however, I was off on exactly what that relationship is. Overall though this is a great guess, especially in somewhat complex symbolic relationships like this.
What happens to the Yo-Yo at the bottom of the string?
Ahh finally a conceptual problem. Here is where building up your physics intuition really comes into play.
So for my guess I’m imagining what’s happening when the Yo-Yo hits the bottom. Well, the string stops unraveling, which means that the Yo-Yo stops spinning around it’s center of mass, and that motion now goes somewhere else. And actually when you think about playing with a Yo-Yo, part of the fun is that you can keep it going – up and down and up and down – over and over again because it is unraveling down, but then re-wrapping the string around the axle on the way up.
Diagram & Variables
Here’s where it’s good to get a little creative and describe the situation as accurately as you can visually.
So first I’m drawing the absolute last point at which the Yo-Yo string has fully unravelled (the finishing point of part (b)). And so because the string becomes fixed and no longer extends downwards with the translation of the Yo-Yo, the Yo-Yo can’t spin around it’s center anymore.
That means that all of that rotational motion is now being transferred to rotation around a fixed point – the “anchor point” at which the string attaches. It then kind of behaves like a pendulum with length b, rotating the center of mass (COM) of the Yo-Yo around that anchor point counter-clockwise until the COM is positioned directly to the left of the anchor point.
Free Body/Torque Diagram & Equations
To get an even deeper insight on this, I’ve again put together a Free Body Diagram (this time combining it with the Torque Diagram), to better understand what’s happening.
Interestingly, now the anchor point of the string is no longer accelerating downwards, so the tension force Ft is now equal to the force due to gravity mg. This means that the center of mass (COM) of the Yo-Yo now accelerates at that instant with a linear acceleration of g (downwards) and an angular acceleration of g/b counter-clockwise around the anchor point.
Okay so now I think we have a better picture of exactly what’s going on with the motion and forces/torques involved at the end of the string. So to answer this question, I’m writing out a description of what happens step-by-step:
- The string anchor point stops moving downwards and becomes fixed.
- The center of mass (COM) of the Yo-Yo accelerates up to g and begins to rotate around the now fixed string anchor point.
- The COM continues accelerating downwards until it is directly below the string anchor point. Here the angular velocity w is at its max.
- The COM continues to rotate around the anchor until it is directly to the left of the anchor, at which point the string begins to wrap around the axle again.
And because this was a pretty relatable physical scenario, my guess was pretty damn close (although less detailed because I went through the process of diagramming everything out).
Okay so thats it for this problem. Hopefully this has helped clarify some of the interactions that go on between linear and rotational motion, forces/torques, and energy.