Rotational motion

[sv3]

This is one area of physics that I still have not mastered. I've only come across these problems occasionally and my prep books don't seem to cover them - unless of course I'm missing the entire concept. How can you take rotational motion and convert it into translational motion? For instance, if you have a merry go-round, and its turning at 10 RPM and you know the diameter - lets say 1 metre, whats the linear velocity? How do you figure the rotational and linear accelerations?

Can you use centripital acceleration formulas on these problems?
I also read on another post on this forum that angular velocity x radius = linear velocity but is it that simple a calcuation?

Anyways, as you can see I'm lost here. I score relatively well in physics 12-13 but the rare time i see these problems, i just blindly guess and keep moving as I just don't know how to handle them Any help is appreciated.

cheers
sv3

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[PhoenixBlaze500]

For instance, if you have a merry go-round, and its turning at 10 RPM and you know the diameter - lets say 1 metre, whats the linear velocity? How do you figure the rotational and linear accelerations?

Disclaimer: I am not absolutely sure about this but I think:

1)To find the linear velocity, you need to find the circumference (C = 2*pi*r; pi = 3.14 and r = radius). 2) Then, multiply the circumference by 10 revolutions per minute to find the linear velocity (v) in meter per minute. 3) If the answers are in meter per second, then you would have to divide what you got in part (2) by 60 (since 1 minute is 60 seconds) to get linear velocity in meter per second.

Is rotational acceleration the same as centripetal acceleration? If so, then a(rotational) = v^2/r

Since the merry-go-round is going in a circular path, I think the linear acceleration would be zero since acceleration is the change in velocity over time and velocity is dependent on displacement. Since displacement is zero (since the merry-go-round is returning to the same point), velocity would be zero, and thus linear acceleration (I think) would be zero. If zero is not one of the choices, then divide the change in linear velocity by time to get linear acceleration.

 

[wanderer]

Disclaimer: I am not absolutely sure about this but I think:

1)To find the linear velocity, you need to find the circumference (C = 2*pi*r; pi = 3.14 and r = radius). 2) Then, multiply the circumference by 10 revolutions per minute to find the linear velocity (v) in meter per minute. 3) If the answers are in meter per second, then you would have to divide what you got in part (2) by 60 (since 1 minute is 60 seconds) to get linear velocity in meter per second.

Is rotational acceleration the same as centripetal acceleration? If so, then a(rotational) = v^2/r

Since the merry-go-round is going in a circular path, I think the linear acceleration would be zero since acceleration is the change in velocity over time and velocity is dependent on displacement. Since displacement is zero (since the merry-go-round is returning to the same point), velocity would be zero, and thus linear acceleration (I think) would be zero. If zero is not one of the choices, then divide the change in linear velocity by time to get linear acceleration.

Rotational acceleration != Centripetal acceleration. alpha = dw / dt
also v = r * w, why would one use circumference?

 

[Isoprop]

This is one area of physics that I still have not mastered. I've only come across these problems occasionally and my prep books don't seem to cover them - unless of course I'm missing the entire concept. How can you take rotational motion and convert it into translational motion? For instance, if you have a merry go-round, and its turning at 10 RPM and you know the diameter - lets say 1 metre, whats the linear velocity? How do you figure the rotational and linear accelerations?

Can you use centripital acceleration formulas on these problems?
I also read on another post on this forum that angular velocity x radius = linear velocity but is it that simple a calcuation?

Anyways, as you can see I'm lost here. I score relatively well in physics 12-13 but the rare time i see these problems, i just blindly guess and keep moving as I just don't know how to handle them Any help is appreciated.

cheers
sv3

Yes, you can use the centrip accel equations if it is satisfying the conditions (constant magnitude of acceleration towards the center, speed is constant but direction of velocity changing, etc.). It should hold true.

Disclaimer: I am not absolutely sure about this but I think:

1)To find the linear velocity, you need to find the circumference (C = 2*pi*r; pi = 3.14 and r = radius). 2) Then, multiply the circumference by 10 revolutions per minute to find the linear velocity (v) in meter per minute. 3) If the answers are in meter per second, then you would have to divide what you got in part (2) by 60 (since 1 minute is 60 seconds) to get linear velocity in meter per second.

Is rotational acceleration the same as centripetal acceleration? If so, then a(rotational) = v^2/r

Since the merry-go-round is going in a circular path, I think the linear acceleration would be zero since acceleration is the change in velocity over time and velocity is dependent on displacement. Since displacement is zero (since the merry-go-round is returning to the same point), velocity would be zero, and thus linear acceleration (I think) would be zero. If zero is not one of the choices, then divide the change in linear velocity by time to get linear acceleration.

Rotational (angular) acceleration is not exactly the same as centripetal acceleration. Angular acceleration is defined as the change in angular velocity per time. It's units are radians/s^s. Centrip acceleration is the center seeking acceleration of an object in uniform circular motion. Its units are in m/s^2.

Example: object is going around uniformly in a circle with a radius of 2 meters. Rotational acceleration is zero because there is no change in angular velocity. But centripetal acceleration is not zero.

Finally, something that has zero displacement doesn't mean it has zero acceleration. Flip a coin. When it comes back to your hand, it has zero displacement. But acceleration has been constant (9.8 m/s^2).

 

[PhoenixBlaze500]

Yes, you can use the centrip accel equations if it is satisfying the conditions (constant magnitude of acceleration towards the center, speed is constant but direction of velocity changing, etc.). It should hold true.



Rotational (angular) acceleration is not exactly the same as centripetal acceleration. Angular acceleration is defined as the change in angular velocity per time. It's units are radians/s^s. Centrip acceleration is the center seeking acceleration of an object in uniform circular motion. Its units are in m/s^2.

Example: object is going around uniformly in a circle with a radius of 2 meters. Rotational acceleration is zero because there is no change in angular velocity. But centripetal acceleration is not zero.

Finally, something that has zero displacement doesn't mean it has zero acceleration. Flip a coin. When it comes back to your hand, it has zero displacement. But acceleration has been constant (9.8 m/s^2).

Thanks for clearing up the rotational acceleration and centripetal acceleration confusion. For the coin example, the acceleration going down would be 9.8 m/s^2, but going up, wouldn't the acceleration be different since the coin is going against gravity?

 

[Isoprop]

Thanks for clearing up the rotational acceleration and centripetal acceleration confusion. For the coin example, the acceleration going down would be 9.8 m/s^2, but going up, wouldn't the acceleration be different since the coin is going against gravity?

Very common misconception.

The entire time the coin is in the air, the acceleration is always 9.8 m/s^2 towards the earth. The moment the coin leaves your hand, it's slowing down at a rate of 9.8 m/s^2 until it reaches the top of its path. Then it's speeding up towards the earth at a rate of 9.8 m/s^2. So the entire time, acceleration is constant.

 

[DJDakoo]

what are the chances of getting a rotational acceleration passage on the real mcat?

 

[sv3]

Yes, you can use the centrip accel equations if it is satisfying the conditions (constant magnitude of acceleration towards the center, speed is constant but direction of velocity changing, etc.). It should hold true.

Rotational (angular) acceleration is not exactly the same as centripetal acceleration. Angular acceleration is defined as the change in angular velocity per time. It's units are radians/s^s. Centrip acceleration is the center seeking acceleration of an object in uniform circular motion. Its units are in m/s^2.

Example: object is going around uniformly in a circle with a radius of 2 meters. Rotational acceleration is zero because there is no change in angular velocity. But centripetal acceleration is not zero.


Thanks for your answers. I am still confused unfortunately. Did I title this thread incorrectly? For example, if you have a merry go-round that is spinning, let's say at a constant 10 RPM, is that actually centripital acceleration and not rotational motion? Perhaps I have these concepts confused? I guess that would be a good start to clearing this up. Because after i figure that out, i want to be able to convert centripital or rotational velocity over to linear velocity. And then finally i can get acceleration info after that. Apologies for not grasping your explanation off the bat.......

sv3

 

[sv3]

Rotational acceleration != Centripetal acceleration. alpha = dw / dt
also v = r * w, why would one use circumference?

So if v = r * w is true, that's great to know! But how do you know w? If something is spinning at 10 RPM, do you just take the circumference and multiply by 10 since thats the total distance per unit of time?

thanks
sv3

 

[Isoprop]

Thanks for your answers. I am still confused unfortunately. Did I title this thread incorrectly? For example, if you have a merry go-round that is spinning, let's say at a constant 10 RPM, is that actually centripital acceleration and not rotational motion? Perhaps I have these concepts confused? I guess that would be a good start to clearing this up. Because after i figure that out, i want to be able to convert centripital or rotational velocity over to linear velocity. And then finally i can get acceleration info after that. Apologies for not grasping your explanation off the bat.......

sv3

Sure, let's clear this up. Centripital acceleration occurs in uniform circular motion. The speed of the object is not changing, only the direction of motion is changing. In other words, its angular velocity is constant.

(I'm using speed here in the strictest sense. Speed is the magnitude of velocity. Speed is a scalar, velocity is a vector.)

Something going at a constant 10 RPM means that it is accelerating but keeping a constant speed. The magnitude of acceleration is constant, and it is undergoing uniform circular motion. So, in your words, it is undergoing centripital acceleration AND rotational motion. But it is not undergoing angular (rotational) acceleration.

Something undergoing angular acceleration is not uniform circular motion. One example would be the merry-go-round that is speeding up. It's giong at 10 RPMs, then 11 RPMs, then 12, etc.

In this merry-go-round, it is still experiencing centripital acceleration, but the magnitude of acceleration is not constant.

Let's sum this up:
centripital acceleration: expressed in m/s^2, is the acceleration experienced by a object going in a circle.
uniform circular motion: magnitude of centriptal acceleration is constant, speed is constant, only direction of velocity is changing. the angular acceleration is zero.
angular acceleration: the thing that is rotating or going in a circle is speeding up or slowing down.

This is pretty hard to explain without drawing it out.

 

[Isoprop]

So if v = r * w is true, that's great to know! But how do you know w? If something is spinning at 10 RPM, do you just take the circumference and multiply by 10 since thats the total distance per unit of time?

thanks
sv3

w is the angular velocity or how many radians does it turn per second. For something going at 10 RPMs, it is making 10 circles per 60 seconds. One circle is 2*pi radians. That means it is going 10*2*pi radians per 60 seconds.

 

[sv3]

Sure, let's clear this up. Centripital acceleration occurs in uniform circular motion. The speed of the object is not changing, only the direction of motion is changing. In other words, its angular velocity is constant.

(I'm using speed here in the strictest sense. Speed is the magnitude of velocity. Speed is a scalar, velocity is a vector.)

Something going at a constant 10 RPM means that it is accelerating but keeping a constant speed. The magnitude of acceleration is constant, and it is undergoing uniform circular motion. So, in your words, it is undergoing centripital acceleration AND rotational motion. But it is not undergoing angular (rotational) acceleration.

Something undergoing angular acceleration is not uniform circular motion. One example would be the merry-go-round that is speeding up. It's giong at 10 RPMs, then 11 RPMs, then 12, etc.

In this merry-go-round, it is still experiencing centripital acceleration, but the magnitude of acceleration is not constant.

Let's sum this up:
centripital acceleration: expressed in m/s^2, is the acceleration experienced by a object going in a circle.
uniform circular motion: magnitude of centriptal acceleration is constant, speed is constant, only direction of velocity is changing. the angular acceleration is zero.
angular acceleration: the thing that is rotating or going in a circle is speeding up or slowing down.

This is pretty hard to explain without drawing it out.

kick asss explanation! Thanks very much. Looks like it was a centripital acceleration/uniform circular motion question and i just didnt recognize it. I had done a similar problem not too long ago where in fact it was angular acceleration and i just hoped that i'd never see it again. Unfortunately i thought thats what this simple problem was. Won't happen again though - thanks for the help!