Converting Between Linear & Rotational - Online Tutor, Practice Problems & Exam Prep

Linear displacement (Δx) relates to angular displacement (Δθ) through the equation ${\mathrm{\Delta x}}^r=\mathrm{\Delta \theta }$. Tangential velocity (v_t) connects to angular velocity (ω) via $v^t=r\omega$. All points on a rigid body share the same angular quantities (Δθ, ω, α), while linear speeds vary with radial distance (r). Understanding these relationships is crucial for analyzing rotational motion and acceleration types, including tangential acceleration (a_t) and centripetal acceleration (a_c).

Converting Between Linear & Rotational

Hey, guys. So you may remember that one of the very first things I showed you in rotation is how we can connect linear displacement, \( \Delta x \), and angular or rotational displacement, \( \Delta \theta \), using a tiny equation. Well, there are 2 more equations that we can use to connect velocity and acceleration between linear and rotational. Okay. So let's check it out.

Alright. So we have these tiny equations that are going to link, that are going to connect, that are going to allow us to convert from one to the other between linear and rotational. Now linear, we're also going to refer to linear as tangential. Right? Linear, tangential, both of these are going in a straight line. Going to connect from linear or tangential to rotational, which is also referred to as angular. So it's important that you know that these words are, have to mean the same thing. Okay. So the linear variable is \( x \) and the rotation equivalent is \( \Delta \theta \). Okay. Now from that, we get that \( \Delta x \) is the change in position, the change in \( x \), and \( \Delta \theta \) is a change in angular or rotational position \( \theta \). And the way that \( \Delta x \) and \( \Delta \theta \) connect is by this equation right here. We've used this. Similarly, \( v \) connects to its angular equivalent, \( \omega \), using a very similar equation. So \( \Delta x = r \Delta \theta \) and \( v \text{ tangential} \) (this \( t \) here means tangential velocity) is \( r \omega \).

And I want to point out that there's a pattern here. This is the linear, the linear variable \( r \) and the rotational variable. The same thing here, linear variable \( r \) rotational variable. I'm going to remove all these little circles so it's not messy. The same thing with \( a \) and \( \alpha \). \( A \) is going to be \( r \), and if you see the pattern, the equivalent of \( v \) is \( \omega \), the equivalent of \( a \) is \( \alpha \). Okay. And this is also the tangential acceleration. So these are the 2 new equations that we're going to be able to use.

Now when do they come up? Usually, it's on a problem like this. You have a disc and this disc spins with angular speed \( \omega \). Well, if you pick a point in this disc, right, a point here, and I want to know what is the velocity, the linear velocity of this point. Well, this point moves with linear velocity or tangential velocity that looks like this, \( dt \). You might remember that when I have a point going around the circle, the point has tangential velocity and it also has centripetal acceleration. Well, it turns out that \( v_t \) is connected to \( \omega \) by this equation, \( v_t = r \omega \), which is a very, very useful relationship equation. Okay? So let's keep going. I want to quickly mention that there are 4 types of acceleration. I already mentioned 2 here. We have \( a_c \) and actually, I already mentioned 3. We have \( a_c \), we have \( a_t \), and we have \( \alpha \). There's a fourth one, but we're going to talk about that later. I want to just be very clear here that this equation right here, \( a_t = r \alpha \) refers to the tangential acceleration. It doesn't refer to the centripetal acceleration. It doesn't refer to the angular or rotational acceleration.

So there are 4 types of acceleration. Most of them have two names. So it's going to be a mess but I'll show you pretty soon, okay. A few more points here. Whenever you have a rigid body or a shape, so let's say this is a cylinder, right, let's say this is a cylinder that spins around itself, okay. All rotational quantities, \( \Delta \theta \), \( \omega \), and \( \alpha \), are the same at every point. So let me show you this real quick, illustrate this a little bit. So let's imagine a line here and then there's point, there's a little imagine this is a huge disc and there's people on top of it or whatever, right? So you have a guy a over here on that point and guy b is over here. Now imagine that this disc spins from here to here. Okay. To that point right there. Now guy a is going to be here and guy b is going to be here. Notice how they all spin on the same line, right? So if I'm here and you're here and this spins, we're still in the same place, right? We're moving together. Okay. So our \( \Delta \theta \), our change in angle will be the same, alright. Because and by the way, this happens even if we're not in the same line. It's just easier to see if it's in the same line. \( \Delta \theta \) is the same. And because \( \omega \) is defined in terms of \( \Delta \theta \), \( \frac{\Delta \theta}{\Delta t} \), \( \omega \) is also going to be the same. And since \( \omega \) is the same, \( \alpha \) depends on \( \omega \), all these three things are the same. Okay? Long story short, if you're in a circle, all the objects on top of a circle have the same \( \Delta \theta \) as they move. They're going to experience the same \( \alpha \) and the same \( \omega \). So all of the rotational quantities will be the same. Okay? However, the linear speeds might be different since they depend on \( r \) which is radial distance, okay, or distance to the center. That's another way to think about it, okay. They might be different. So the best way to illustrate this is by doing an example and do a very straightforward one. So I have a wheel of radius 8. So let's draw this here. Just put a little radius here. Radius of this wheel is 8 meters. It spins around its central axis. So what that means is that imagine a circle and imagine a sort of an invisible line through the circle, right. An invisible line through the circle and it's free to spin around that invisible line. So I'm going to draw this here. You don't have to draw it. So I'm going to delete it. Imagine the imaginary line that goes through this thing, almost as if you stuck a thing through it and then it's free to spin around that. Okay. That's what that means. Let's get this out of here. Basically, it spins around at center, which is how these things always work, at 10 radians per second. So that's our \( \omega \) is going to be, 10 radians per second. We want to know the angular and linear speeds at different points. So I want to know at a point in the middle of the wheel on the central axis, so. So we're going to call this point 1 at a distance 4 meters from the center. If the radius is 8 meters, 4 meters is halfway in. I'm going to draw this here. This is point 2 and at the edge of the wheel, point 3. Okay? So what we want to know is we want to know \( V_1 \), \( V_2 \), and \( V_3 \). And and I want to know \( \omega_1 \), \( \omega_2 \) and \( \omega_3 \). Okay? That's what it says. I want the angular, which is \( \omega \), and linear, \( v \), speeds at these three points. K? So first thing is to realize that all these points have the same because they're on the same disk, they have the same \( \omega \) and that \( \omega \) is the same \( \omega \) as the disk. Okay? So that's the first part. \( \omega_1 = \omega_2 = \omega_3 = \omega_{\text{disk}} \). So this is more of a conceptual to know that you to know if you know that. So all of these will be 10

Length of string rotating point mass

Alright. So here we have a small object that rotates at the end of a light string. So here's a string, and you got a small object. I'm going to do this, and it's spinning like this. Okay? Imagine that, if you spin a string, it forms a circular path. Right? The object reaches 120 RPM from rest in just 4 seconds. I'm giving you a ton of information here. First, you start at rest, so make the initial 0. You reach the final RPM of 120, and you did this in just 4 seconds. It also says here that the tangential acceleration, the tangential acceleration remember is \( a_t \), after the 4 seconds is, 15 meters per second squared. And we want to know what is the length of the string. Okay. We haven't talked about the length of the string yet, but I hope you can figure out that the length of the string is the distance here. Right? It's the radius of the circle that forms the circular path you get, or it is the radial distance from the center of rotation, which is your hand, and the edge of rotation, which is where the object is. So little \( r \), the distance to the middle, is your length. So essentially, what we're looking for is little \( r \). Think of it as little \( r \), not as \( l \) because there's no \( l \) in any of these equations. So you're not going to find \( l \). Okay. So this is a little bit of a mess because we have to use a combination of equations here. Okay? So if you look through all the equations we've used so far, you might first think about 1 of the 3 or 4 motion equations, 1, 3, or 4 motion equations. And you might think of that because I gave you \( \omega \) initial is 0. I gave you \( \Delta t \). I gave you RPM which you can convert to \( \omega \) final. And if you do that, you're going to have 3 out of 5 variables once you convert. Okay? However, notice that if you look through all those four equations, there are no \( r \)'s in them. So you're not going to be able to solve for \( r \) by doing this. Okay? Now if you look a little further, I do have an equation that I gave you recently that links up \( a_t \) with \( r \). And that equation is \( a_t = r \alpha \). And I know \( a_t \), so all I have to do is find \( \alpha \). \( r \) equals \( a_t \), which is 15 right there, divided by \( \alpha \). So what we're going to have to do is find \( \alpha \) and plug it in here. Now how do I find \( \alpha \)? Well, \( \alpha \) is one of my five variables of motion, so I'm going to be able to use these 3 guys to find \( \alpha \). Okay? And that's what we're going to do now. So first, I'm going to convert from RPM into frequency. I'm sorry, into \( \omega \) final. So remember that \( \omega \) final is \( 2\pi f \) and \( f \) is RPM over 60. So 120 divided by 60 is 2. Therefore, \( \omega \) final is \( 2\pi \) and instead of \( f \), I'm going to put a 2, which is \( 4\pi \). Okay? So I'm going to rewrite this here just to clean it up. \( \omega \) initial is 0. \( \omega \) final is \( 4\pi \). \( \Delta T \) is 4. We're looking for \( \alpha \) and the ignored variable is \( \Delta \theta \). Notice how I know three things. The ignored variable is \( \Delta \theta \). The only equation that doesn't have \( \Delta \theta \) is the first one out of the 4, \( \omega \) final equals \( \omega \) initial plus \( \Delta e \). And if we're looking for \( \alpha \), we just got to move everything out of the way. \( \omega \) initial is 0. So \( \alpha \) is \( \omega \) final divided by time. \( \omega \) final is, we found here, it's \( 4\pi \). Divided by time, which is 4. \( \alpha \) is \( \pi \) radians per second squared. Now all we got to do is plug in this number here, and we're good. So 15 divided by \( \pi \). And if you divide the 2, you get that \( r \) is 4.77 meters. Okay. 4.77 meters, and that is the final answer. Cool. So that's it for this one. It's an interesting question that combines these two equations. And the basic idea here is just that old school physics hustle of, you got stuck in one, then you're going to have to go find the other and just kind of work your way through it. Alright? There's not a very clear path. There are a few different ways you could have done this. But the most important thing is, you know, try to figure out what's an equation that has my variable and then sort of look for all, different ways to find all the variables you have to solve for. Cool. Alright. That's it for this one. Let me know if you guys have any questions.

A disc of radius 10 m rotates around itself with a constant 180 RPM. Calculate the linear speed at a point 7 m from the center of the disc.

• A. 18.8 m/s
• B. 21.0 m/s
• C. 66.0 m/s
• D. 132 m/s

A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds. Calculate its tangential acceleration after 3 s. 

BONUS:Calculate its tangential speed after 3 s.

• A. 1.31 m/s²
• B. 15.8 m/s²
• C. 20.9 m/s²
• D. 62.7 m/s²

A 4 m long blade initially at rest begins to spin with 3 rad/s² around its axis, which is located at the middle of the blade. It accelerates for 10 s. Find the tangential speed of a point at the tip of the blade 10 s after it starts rotating.

• A. 0 m/s
• B. 15 m/s
• C. 30 m/s
• D. 60 m/s
• E. 188 m/s

Here’s what students ask on this topic:

How do you convert linear displacement to angular displacement?

Linear displacement (Δx) can be converted to angular displacement (Δθ) using the equation:

$\mathrm{\Delta x}=r\mathrm{\Delta \theta }$

Here, Δx is the linear displacement, r is the radius or distance from the center of rotation, and Δθ is the angular displacement. This relationship is crucial for understanding how a point on a rotating object moves in a straight line as the object rotates.

What is the relationship between tangential velocity and angular velocity?

Tangential velocity (v_t) is related to angular velocity (ω) through the equation:

$v$_t=rω

In this equation, v_t is the tangential velocity, r is the radius or distance from the center of rotation, and ω is the angular velocity. This relationship helps in determining the linear speed of a point on a rotating object.

How do you calculate tangential acceleration from angular acceleration?

Tangential acceleration (a_t) can be calculated from angular acceleration (α) using the equation:

$a$_t=rα

Here, a_t is the tangential acceleration, r is the radius or distance from the center of rotation, and α is the angular acceleration. This equation is useful for understanding how the linear acceleration of a point on a rotating object changes as the object accelerates angularly.

Why do all points on a rigid body have the same angular quantities?

All points on a rigid body have the same angular quantities (Δθ, ω, α) because they rotate together as a single unit. This means that the angular displacement (Δθ), angular velocity (ω), and angular acceleration (α) are uniform across the entire body. However, the linear speeds of different points vary with their radial distance (r) from the center of rotation. This uniformity in angular quantities is essential for analyzing rotational motion and ensuring consistent behavior across the rigid body.

How does radial distance affect linear speed in rotational motion?

In rotational motion, the linear speed (v) of a point on a rotating object depends on its radial distance (r) from the center of rotation. The relationship is given by:

$v=r\omega$

Here, v is the linear speed, r is the radial distance, and ω is the angular velocity. Points farther from the center (larger r) have higher linear speeds, while points closer to the center (smaller r) have lower linear speeds. This is why objects at the edge of a rotating disk move faster than those near the center.