Types of Acceleration in Rotation - Online Tutor, Practice Problems & Exam Prep

In rotational motion, four types of acceleration are crucial: centripetal acceleration (A_C), tangential acceleration (A_T), total acceleration (A), and angular acceleration (α). A_C maintains circular paths, while A_T and α indicate changes in speed. The relationship A = √(A_T² + A_C²) combines these accelerations. For example, in a carousel, if the angular velocity is constant, A_T and α are zero, simplifying total acceleration to A_C. Understanding these concepts is essential for solving rotation problems effectively.

Types of Acceleration in Rotation

Hey, guys. So now we're going to talk about the 4 different types of accelerations you'll see in rotation problems. It can get a little overwhelming because it's a lot of letters, a lot of variables, a lot of equations, but I'm going to try to simplify this as much as possible. Let's check it out.

Alright. So, there are 4 types of acceleration in rotation problems as I said. Most of them have two names. So in total, if you look, you have 7 different names and you have to know how the pairs which pair of names go together. Centripetal acceleration, a_c, also referred to as radial acceleration, a_rad. Those are the same thing. Tangential acceleration is a_t. Sometimes, this is referred to as linear acceleration. The other one is total acceleration which sometimes is referred to either as total acceleration or just acceleration. Sometimes it's referred to as the total linear acceleration. So this is just a. It's a total acceleration. And then the last one is rotational or angular acceleration. You used this one. This is α. I want to point out that these guys are measured in meters per second squared and this is measured in radians per second squared. All these 3 here, these 3 are linear accelerations, so they're represented with sort of a straight line as opposed to alpha, which is an angular acceleration. So it's measured this way.

Now there are these 4 types, but depending on what kind of situation you have, you're not going to have all 4 types. And what that means is that some of them will just be 0. So first, you always have v_t and a_c. So if you're going around a circle, that little object that's going around the circle, let's say this way, at this point, this object has a tangential velocity and it has a centripetal acceleration pointing towards the middle. It's spinning this way with Ω. Remember that at this point, v_t equals rΩ. Remember also that a_c equals v squared over r. This is from a while ago. And what I can do is I can actually rewrite v as rΩ. So it's going to look like this, rΩ squared over r. So it's going to be like this, rΩ squared. These always exist. You always have a_c because a_c, if you remember from F equals ma, you have a_c because you have a centripetal force. There's a force that pulls you towards the middle. Therefore, you have a centripetal acceleration. a_c is responsible for maintaining the circular path.

Another way to think about this is that it keeps the object spinning. If a_c doesn't exist, you can think of it as there's no longer a force no longer a force pulling into the middle. So this object is going to go straight up, in this direction. So a_c, it's there and it maintains a circular path. So as long as you're moving a circle, you have to have a_c and you have to have v_t. And you have an Ω because obviously you're spinning. Now, if you're accelerating, if you're accelerating, you are also going to have a_t and α. The converse of that is that if you're not accelerating, these guys are 0.

So what does a_t look like? The idea is that you're not just spinning in place, but at this point, the object is actually getting faster. So if this thing is rotating like this, you can think of it that to get faster, you have to like something has to push the object this way, some force. And if the object is going faster this way, it's also rotating faster, so there's also an α this way. So you got an Ω and you got an α in that same direction.

Now remember, you always have your a_c. So this sets up an interesting situation where you have a_t this way and a_c this way. They make an angle of 90 degrees with each other right there. What happens is you have 2 arrows this way, so you can combine the two of them using vector composition. And this is what we call the total acceleration, sometimes referred as the total linear acceleration, or sometimes referred simply as acceleration.

So if you don't see if you see just acceleration without the words centripetal, radial, tangential, linear, rotational, angular, if you don't see any of these words, you can assume it's just regular acceleration, the total acceleration, which is a combination of these 2. It's just vector addition. So a is the square root, it's the Pythagorean of the two sides. So a_t squared plus a_c squared.

So that's what the total acceleration. So whereas a_c maintains your circular path, these guys here are responsible for changing angular speed. These guys, a_t changes angular velocity. Now changes means it could be going faster or slower.

This equation, a_tα, which is one that I gave you earlier, is a good way to remember that a_t and α are connected. You have this and this or they're both 0. These guys are both either nonzero together or they're both 0 together. And then the last point I want to make is that if your a_t is in fact 0, which would mean that α is 0, look at this equation right here, then your total a, which is usually the square root of a_t squared plus a_c squared simplifies. If your a_t is 0, look what happens. You get simply 0 plus a_c squared, which is the square root of a_c squared, which is simply a_c.

I'm just pointing that out in case you see that, and you don't think it's weird that there's 2 a's inside of the square root, but one goes away, and then your a becomes just your a_c. That's totally doable. In fact, that's what happens here. Here, your a is only a_c. So it's like you have these two arrows stacked on top of each other.

Four accelerations, and I gave you 2 new equations, one new way to rewrite a_c and an equation for the total a. Alright? So let's do a problem here. It's got 5 parts. It's kind of annoying, but hopefully you see that it's not that bad. I have a carousel, a carousel 10 meters in radius. So a big circle radius equals 10 meters. I know I don't have a lot of space here, but you actually don't need that much room. But let's write small

.

Complete one cycle. One cycle every 75 seconds. So that's the period. T is 45 seconds. I want to know what is the tangential velocity. There's a boy that stands at the edge. So if the boy is at the edge, the boy is at a distance r equals 10. This thing is 10 meters long in radius. If you're sitting at the edge, you sit at the 10-meter distance from the center. And I want to know his tangential velocity. So tangential velocity is v_t. And remember, the tangential velocity of a point, a person, an object, whatever, on a circle, on any kind of spinning object is rΩ, where Ω is the Ω of the boy, which is the same as the Ω of the disc.

So the boy is at the distance 10, but I don't know Ω. However, remember, I can get Ω from T because Ω, frequency, T, and rpm are all interconnected. Ω is 2π over T. So 2π divided by 45 seconds. And if you do that, you get 0.14. And that's what goes here, 0.14. Therefore, the answer is 1.4 meters per second is the velocity that the boy experiences.

For part b, I want to know what is the angular acceleration. Angular acceleration is α. Now remember, you only have α if you're actually speeding faster. Once it says that it completes one cycle every 45 seconds, right, first 45 seconds, one cycle, second 45 seconds, another cycle, It implies that this is a constant, it's a constant movement at a constant rate, at a constant velocity. So α is actually 0.

Radial acceleration is a_rad which is the same as a_c which is v squared over r or rΩ, whichever you prefer. So I'm going to use rΩ, just so I don't have to square this number, but it's the same exact thing. R is 10, and Ω is points I'm sorry. It's rΩ squared. I was like, hey, that's the same thing as that. So I'm going to have to square something either way, 0.14 squared. But anyway, if we do this, we get that the answer is 0.196 meters per second.

For part d, keeping it tight there, the tangential acceleration is a_t. Remember, a_t only exists if you're actually pushing this thing to spin faster. Another way to think of this is that a_t is rα and α 0, therefore, a_t equals 0.

So part of the reason why this question doesn't require that much space is because some of the answer is 0. The total linear acceleration is a, which is the square root of a_t squared plus a_c squared, And a_t is 0, so you're left with the square root of a_c, which is just a_c, which is the same thing as a_rad, which is 0.196 meters per second squared. So here's the 5 answers. This, 0, this, 0. And the total a right here is that. Alright? So just a bunch of equations and knowing how to link everything together. It's good to do some practice and make sure you know how to do this. It's pretty straightforward but it's kind of annoying. Alright? So hopefully, this makes sense. Let me know if you guys have any questions.

Accelerations of boy on carousel

Alright. So in the second example, we have a carousel with a radius of 16 meters. Let's illustrate a circle where the radius is 16 meters. It accelerates from rest, with the initial angular speed, ω_initial, being 0 and angular acceleration, α, being 0.05. A buoy stands at the edge, so the little r, the distance between the center and the buoy, is 16 meters. We want to know, after 10 seconds (Δt = 10 seconds), what his tangential velocity is.

I'm going to organize this information on the side because tangential velocity v_t is not one of the five motion equations. Recall that the tangential velocity or linear velocity of an object at a point in the disk is given by rω, where r is the distance from the center and ω is the angular speed. We know r = 16, but we don't have ω. So, we need to find ω_final using motion equations. The variable that's missing here is Δθ, so I'll put a little sad face here. This tells me that the equation I should use is the first one, because it's the only equation that doesn't require Δθ. The equation is ω_final = ω_initial + αt. We're looking for ω_final. Since ω_initial = 0, α = 0.05, and t = 10, ω_final = 0 + 0.05 * 10 = 0.5 radians per second. Plugging this in, we get v_t = 16 * 0.5, which gives us a velocity of 8 meters per second.

For Part B, we want to know the tangential acceleration, a_t. This is calculated as rα. With r = 16 and α = 0.05, we get a_t = 16 * 0.05 = 0.8 meters per second squared.

For Part C, we are looking for the radial acceleration, a_rad, which is the same as centripetal acceleration, a_c. We can use v² / r or rω². I'll use the second formula since I already computed v_t. It's going to be (8²) / 16 = 64 / 16 = 4 meters per second squared.

For Part D, we are interested in the angular acceleration. Angular acceleration is simply α. We already have the angular acceleration, so it's going to be 0.05 radians per second squared.

Finally, the total linear acceleration, a, is a combination of the other two linear accelerations. It's calculated as √(a_t² + a_c²). If you plug this in, it's √(0.8² + 4²) = √(0.64 + 16) = √(16.64) ≈ 4.1 meters per second squared. These are the final answers.

Alright. That's it for this stuff. Let me know if you have any questions.

A large disc of radius 10 m initially at rest takes 200 full revolutions to reach 30 RPM. Calculate the total linear acceleration of a point at half way between the disc's center and its edge, once the disc reaches 30 RPM. (You may assume it continues accelerating past that point)

• A. 15.7 m/s²
• B. 24.7 m/s²
• C. 49.3 m/s²
• D. 98.7 m/s²

An object of negligible size moves in a circular path of radius 20 m with 90 RPM. Find its radial acceleration.

• A. 0 m/s²
• B. 444 m/s²
• C. 592 m/s²
• D. 1776 m/s²

Here’s what students ask on this topic:

What are the different types of acceleration in rotational motion?

In rotational motion, there are four key types of acceleration: centripetal acceleration (AC), tangential acceleration (AT), total acceleration (A), and angular acceleration (α). Centripetal acceleration (AC) maintains the circular path of an object and is directed towards the center of the circle. Tangential acceleration (AT) indicates changes in the speed along the tangent to the circular path. Total acceleration (A) is the vector sum of centripetal and tangential accelerations, given by the equation $\sqrt{a^2+a^2}$. Angular acceleration (α) describes the rate of change of angular velocity and is measured in radians per second squared.

How do you calculate centripetal acceleration in rotational motion?

Centripetal acceleration (AC) in rotational motion can be calculated using the formula $ac=\frac{v^2}{r}$, where $v$ is the tangential velocity and $r$ is the radius of the circular path. Alternatively, it can also be expressed as $ac=r{\omega }^2$, where $\omega$ is the angular velocity.

What is the relationship between tangential acceleration and angular acceleration?

Tangential acceleration (AT) and angular acceleration (α) are related by the equation $at=r\alpha$, where $r$ is the radius of the circular path. This relationship shows that tangential acceleration is directly proportional to both the radius and the angular acceleration. If the angular acceleration is zero, the tangential acceleration will also be zero, indicating no change in the tangential speed of the object.

How do you find the total acceleration in rotational motion?

Total acceleration (A) in rotational motion is the vector sum of centripetal acceleration (AC) and tangential acceleration (AT). It can be calculated using the Pythagorean theorem: $A=\sqrt{a^{\mathrm{t}}+a^{\mathrm{c}}}$. This equation combines the magnitudes of the tangential and centripetal accelerations to give the total acceleration experienced by the object in rotational motion.

What is the significance of angular acceleration in rotational motion?

Angular acceleration (α) is significant in rotational motion as it describes the rate of change of angular velocity. It indicates how quickly an object is speeding up or slowing down its rotation. Angular acceleration is measured in radians per second squared and is crucial for understanding the dynamics of rotating systems. It helps in determining the tangential acceleration and, consequently, the total acceleration of the object in rotational motion.