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## Physics library

# Rotational kinematic formulas

David explains the rotational kinematic formulas and does a couple sample problems using them. Created by David SantoPietro.

## Want to join the conversation?

- At14:03, how did the answer become -6.37 m/s^2? I substituted the same values in the exact same kinematics equation and I got -62.83 m/s^2. Did I possibly enter something wrong with my calculator (it was in radians mode).(20 votes)
- I know the answer to your question!

Ok, so when you put them into the calculator, ALWAYS PUT PARENTHESIS FOR PI!

It is so important! If you just divide -(40rad/s)^2 by 80pi (2*40pi), then you will get -62.8319...

If you divide -(40rad/s)^2 by (80pi), you will get the right answer -6.3662...(34 votes)

- At 11 minutes, why did you use the full 4m for R? When working the problem I assumed 4m to be diameter so I thought R would be equal to 2m?(6 votes)
- There is no diameter here. r is the distance from the pivot point to where the force is applied. It's 4 meters.(5 votes)

- At11:34, How is the answer 160m/s ? where did the radians go? Why not 160m rad/ sec? Thanks.(9 votes)
- How can
`(4m)(40rad/s)`

be equal to`160m/s`

? I know the numbers are correct, but units don't really work, it should be`rad×m/s`

.(3 votes)- As it happens, the

can be ignored, because radians are at their heart a ratio. And ratios are unitless, because`rad`

`5 units / 10 units = 1/2 (unitless)`

But you can leave it there if you want, it is still technically correct.

A radian is based on the formula s = r(theta). We use radians because if we plug in s = rx, some multiple of the radius, we cancel r to get x = theta, and since x is just a constant multiple, we have unitlessly defined an angle, which is extremely useful compared to degrees, which are arbitrary and would mess up the formula for this if you were to use them.(6 votes)

- Where does the 4th kinematic formula comes from?

3 months ago by Roma

I have the same question. The 4th kinematic formula is v^2=V^2 +2ax. David showed 4 formulas at about2:26, which he said only apply if acceleration is constant. Where does the last formula come from?(3 votes)- The kinematic equations are a set of four equations that can be utilised to predict unknown information about an object's motion if other information is known. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion.(2 votes)

- what is the good way to memorize the kinematic formulas? I understand how they are used but there are four that looks different from each others...(2 votes)
- Hi, I suggest you think about how you can derive them.

For example, take`v^2 = v0^2+2a delta x`

You see that since`v^2 = (v0 + at)^2`

= v0^2 + 2v0at + a^2t^2

= v0^2 + 2a (v0t + 1/2at^2)

= v0^2 + 2a delta x

If you know some calculus, it can be shown that`delta x = integral (v0 + at) dt`

which makes sense because position is the 'area' under velocity.

Finally, for`delta x = (v + v0) t / 2`

It's really just average velocity times time

Hope that helps!(1 vote)

- What if the angular velocity is not constant? What would the new equations be? How would I derive them?(1 vote)
- These formulas are for constant
**angular acceleration**, not constant angular velocity. If the angular acceleration is not constant, then the only way to solve rotational kinematics questions would be to use calculus -- the equations would involve derivatives and integrals.(3 votes)

- Can these kinematic formulas also be used for the arc length, tangential velocity, and tangential acceleration? Or do you have to do all your calculations with theta, angular velocity, and angular acceleration and then just convert at the end?(2 votes)
- How can (4m)(40rad/s) be equal to 160m/s? I know the numbers are correct, but units don't really work, it should be rad×m/s.(1 vote)
- Think of how a radian is defined? A radian is a distance along the circumference of a circle divided by the radius (a distance from the center of the circle to the circumference). So a radian per second is (m/m)/s = m/(m*s). When you multiply the m/(m*s) by the radius you have (m * m/)(m*s) = m/s.(2 votes)

- 11:25

Where did the radians unit go? Why isn't it 160 (m rad)/s?(1 vote)- Radians are actually m/m where it is a ratio of distance around the circle over the radius of the circle so radians are unitless.(2 votes)

## Video transcript

- [Instructor] So in the
previous couple videos, we defined all these new
rotational motion variables and we defined them exactly
the same way we defined all these linear motion variables. So for instance, this angular displacement was defined the exact same way we defined regular displacement, it's just
this is the angular position as opposed to the position,
the regular position. Similarly this angular velocity was the angular displacement per
time just like velocity was the regular displacement over time. And the angular acceleration
was the change in the angular velocity per time,
just like regular acceleration was the change in regular
velocity per time. And so because these
definitions are exactly the same except for the fact that
the linear motion variable is replaced with its angular counterpart, all the equations results in
principles we found and derived for the linear motion
variables will also hold true for the rotational motions
variables as long as you replace the linear motion variable
in that equation with its rotational motion variable counterpart. And it even works with graphs. So let's say you had a
velocity versus time graph and it looked like this. Since we already know from 1D
motion that the slope of this velocity versus time graph
is equal to the acceleration, that means on an angular
velocity versus time graph, the slope is going to represent
the angular acceleration, because the relationship
between omega and alpha is the same as the
relationship between v and a. Similarly the area underneath the curve on a velocity versus time graph represented the displacement. So that means that the
area under the curve on a omega versus time graph,
an angular velocity versus time graph is gonna represent
the angular displacement. And so if you remember
from 1D motion, the way we derived a lot of the 1D
kinematic formulas that related these linear motion variables,
was by looking for areas under a velocity graph. We could do the same thing for the rotational motion variables. We could find this area,
relate it to omega and alpha and we'd get the rotational
kinematic formulas, but we already know since
these are all defined the same way the linear
motion variables are defined, we're gonna get the exact
same equations, just with the linear motion variable
replaced with its rotational motion variable. So let's right those down. First we'll right down the
linear motion kinematic formulas. If you remember they looked like this. So there they are. These are the four kinematic
formulas that relate the linear motion variables. But remember this only works,
these equations only work if the acceleration is constant. But if the acceleration is constant, these four kinematic formulas
are a convenient way to relate all these kinematic
linear motion variables. Now if you wanted rotational
kinematic formulas, you could go though the
trouble that we went through with these to derive them
using areas under curves, but since we know the
relationship between all these rotational motion variables
is the same as the relationship between the
linear motion variables, I can make rotational
motion kinematic formulas simply by replacing all
of these linear variables with their rotational motion
variable counterparts. So let's do that. So in other words, instead
of V, the velocity, the final velocity, I would have omega, the final angular velocity. Instead of V initial,
the initial velocity, I'd have the initial angular velocity. Instead of acceleration, I'd
have the angular acceleration. And time is just time. So there's no such thing as
angular time or linear time. As far as we know, there's
only one time and that's t and that works in either equation. So you could probably guess,
when are these rotational motion kinematic formulas gonna be true? It's gonna be when the alpha, the angular acceleration is constant. And so you can keep goin' through, wherever you had an x, that
was the regular position, you'd replace it with
theta, the angular position. So I'll replace all these x's with thetas. We replace all of our accelerations with angular accelerations. And then I'll finish cleaning up these v initial and v finals. And then we've got 'em. These are the rotational
kinematic formulas. The are only true if the angular
acceleration is constant, but if it is constant,
these are a convenient way to relate all these
rotational motion variables and you can solve a ton a problems using these rotational kinematic formulas. And in fact, you use
these, the exact same way you used these regular kinematic formulas. You identify the variables that you know. You identify the variable
that you wanna find and you use one of the
formulas that lets you solve for that unknown variable. So let me show you some examples. Let's do a couple examples
using these formulas, cause it takes a while before
you get the swing of 'em. So let me copy these. We're about to use these. And let's tackle a couple examples of rotational kinematic formula problems. So let me get rid of all this
and let's tackle this problem. Let's say you had a four meter long bar, that's why I've had this
bar here the whole time, to show that it can rotate. It starts from rest and
it rotates through five revolutions with a constant
angular acceleration of 30 radians per second squared. And the question is, how
long did it take for this bar to make the five revolutions? So what do we do? How do we tackle these problems? You first identify all the
variables that you know. So it said that it
revolved five revolutions, that's the amount of angle
that it's gone through, but it's in weird units. This is in units for revolutions. So we know what the delta theta is, five revolutions. But we want our delta theta
always to be in radians, cause look it, our acceleration was given in radians per seconds squared. You've gotta make sure you
compare apples to apples. I can't have revolutions for delta theta and radians for acceleration. You've gotta pick one unit to go with and the unit we typically
go with is radians. So how many radians would
five revolutions be? One revolution is two pi
radians, cause one time around the entire circle is two pi radians. That means that five revolutions would be five times two pi radians, which gives us 10 pi radians. So we've got our angular
displacement, what else do we know? It tells us this 30
radians per second squared. That is the angular acceleration. So we know that alpha is 30
radians per second squared. You can write the radian,
you can leave it off. Sometimes people write the radian, sometimes they leave it blank. So you can write one over
second squared if you wanted to. That's why I left this blank over here, but we could write
radians if we wanted to. And should this be alpha
be positive or negative? Well, since this object is speeding up, it started from rest,
that means it sped up. So our direction of the
angular displacement, has to be the same direction
as this angular acceleration. In other words, if
something's speeding up, you have to make sure that
your angular acceleration has the same sign as
your angular velocity, and your angular velocity'll
have the same sign as your angular displacement. So since we called this
positive 10 pi radians and the object sped up, we're
gonna call this positive 30 radians per second squared. If this bar would have slowed
down, we'd of had to make sure that this alpha has the opposite sign as our angular velocity. But that's only two rotational
kinematic variables. You always need three in
order to solve for a fourth. So what's our third
rotational kinematic variable? It's this. That it says the object started from rest. So this is code. This is code word for
omega initial is zero. Initial angular velocity is
zero, cause it starts from rest. So that's what we can say down here. That's our third known variable. And now we can solve. We've got three, we
can solve for a fourth. Which one do we want to solve for? It says how long, so that's the time. We wanna know the time that it took. All right so these are the
variables that are involved. We wanna know the time. We know the top three. The way I figure out what
kinematic formula to use is that I just ask which
variable's left out of all these? I've got my three knowns
and my one unknown that I want to find. Which variable isn't involved? And it's omega final. So omega final is not
involved here at all. So I'm gonna use the
rotational kinematic formula that does not involve omega final. I'll put these over here. So I'll look through 'em. First one's got omega final. I don't wanna use that one cause I wouldn't know what to plug in here and I don't wanna solve for it anyway. I don't want the second one. This third one has no omega final, so I'm gonna use that one. So let's just take this,
we'll put it over here. So we know delta theta. Delta theta was 10 pi radians. And we know omega initial was zero. So this whole term is zero. Zero times t is still
zero, so that's all zero. And we have 1/2. The angular acceleration was 30 and the time is what we wanna know. And you can't for get that that's squared. So now we just solve this
algebraically for time. We multiply both side by two. That would give us 20 pi. Then we divide by 30. And that'll end up giving us 20 pi, and technically that is 20 pi radians divided by 30 radians per second squared and then you have to take the square root, because it's t squared. And if you solve all this for t, I get that the time ended up
taking about 1.45 seconds. And our units all canceled
out the way should here. Radians canceled radians. You ended up with seconds
squared on the top. You too the square root. That gives you seconds to end with. Now this second part, part b, says what was the angular
velocity after rotating for five revolutions? Now there's a couple
ways we could solve this. Because we solved for the
time, we know every variable except for the final angular velocity. So I could use any of these now. To me, this first one's the simplest. There's no squares involved. There's not even a ratio or
anything, so let's use this. We could say that omega
final is gonna equal omega initial, that was just zero, plus the angular acceleration was 30, and now that we know the
time we could say that this time was 1.45 seconds. And that gives me a final angular velocity of 43.5 radian per second. That's how fast this thing
was revolving in a circle the moment it hit five revolutions. So that was one example. Let's do another one. Let's carry our kinematic
formulas with us. We could use those. So we get rid of all that. Let's check this one out. Says this four meter
long bar is gonna start, this time it doesn't start from rest. This time it starts with
an angular velocity, oh, we're not gonna rotate that. Whoa, that'd be a more difficult problem, we're gonna rotate this. This four meter long bar
starts with an angular velocity of 40 radians per second,
but it decelerates to a stop after it rotates 20 revolutions. And the first question is, how
fast is the edge of the bar moving initially in meters per second? So in other words, this
point on the bar right here, is gonna have some velocity this way. We wanna know, what is
that velocity initially in meters per second? Well this isn't too hard. We've got a formula that relates the speed to the angular speed. You just take the distance from the axis, to the point that you want
to determine the speed and then you multiply it
by the angular velocity and that gives you what
the speed of that point is. So this r, let's be careful, this is always from the axis. And in this case this
is the axis right there. The distance from the axis
to the point we wanna find is in fact the entire length of this bar, so this will be four meters. So to find the speed we
could just say that that's equal to four meters, since you wanna know the
speed of a point out here that's four meters from the axis, and we multiply by the angular velocity, which initially was 40 radians per second. And we get the speed of
this point on the rod, four meters away from the
axis is 160 meters per second. That's really fast. And that's the fastest point on this rod. If you were gonna ask
what the speed of the rod would be halfway, that
would be half as much. Because this would only
be, this r right here, would only be two meters
from the axis to that point, it's only two meters. And the closer in you go,
the smaller the r will be, the smaller the speed will be. So these are gonna travel, these
points on the rod down here don't travel very fast at all, because their r is so small. All these points have the
same angular velocity. They're all rotating with the same number of radians per second, but the
actual distance of the circle they're traveling through is different, which makes all of their speeds different. So that answers part a, we got how fast in meters per second. It was going 160 meters per second. And the next part asks, what
was the angular acceleration of the bar? All right, this one we're
gonna have to actually use a kinematic formula for. We'll bring these back, put 'em over here. Again the way you use these,
you identify what you know. We know the initial
angular velocity was 40. So this time we know omega
initial 40 radians per second. Set it revolved 20 revolutions. That's delta theta, but
again, we can't just write 20. We've gotta right this in terms of radians if we're gonna use these
radians per second. They have to all be in the same unit. So it's gonna be 20 revolution times two pi radians per revolution. So that's 40 pi radians. What's our third known? You always need a third known
to use a kinematic formula. It's this. It says it decelerates to a
stop, which means it stops. That means omega final, the
final angular velocity is zero. And we want the angular
acceleration, that's alpha. So this is what we wanna know. We wanna know alpha. We know the rest of these variables. Again to figure out which equation to use, I figure out which one got left out. And that's the time. I was neither give the time nor was I asked to find the time. Since this was left out, I'm
gonna look for the formula that doesn't use time at all. And that's not the first one. That's not the second or the third, it's actually the fourth. So I'm gonna use this fourth equation. So what do we know? We know omega final was zero. So I'm gonna put a zero squared. But zero squared is still zero, equals omega initial squared. That's 40 radians per second squared. And then it's gonna be
plus two times alpha. We don't know alpha, but
that's what we wanna find, so I'm gonna leave that as a variable. And then delta theta we know. Delta theta was 40 pi radians
since it was 20 revolutions. And if you solve this
algebraically for alpha, you move the 40 over to the other side. So you'll subtract it. You get a negative 40
radians per second squared. And then you gotta divide by this two as well as the 40 pi radians, which gives me negative 6.37
radians per second squared. Why is it negative? Well this thing slowed down to a stop. So this angular acceleration
has gotta have the opposite sign to the initial angular velocity. We called this positive 40, that means our alpha's gonna be negative. So recapping, these are the
rotational kinematic formulas that relate the rotational
kinematic variables. They're only true if the angular
acceleration is constant. But when it's constant, you
can identify the three known variables and the one unknown
that you're trying to find and then use the variable
that got left out of the mix to identify which
kinematic formula to use, since you would use the
formula that does not involve that variable that was
neither given nor asked for.