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### Course: High school physics (DEPRECATED)>Unit 7

Lesson 2: Angular kinematics

# 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?

• At , 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).

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...
• 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?
• There is no diameter here. r is the distance from the pivot point to where the force is applied. It's 4 meters.
• At , How is the answer 160m/s ? where did the radians go? Why not 160m rad/ sec? Thanks.
• radians are “unitless.” you can essentially add or drop them wherever you like without changing the value, and they’re very conveniently defined so that you can multiply the radius by radians / second and get speed
• 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`.
• As it happens, the
``rad``
can be ignored, because radians are at their heart a ratio. And ratios are unitless, because
`` 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.
• In the first problem, how the root of 20/30 is 1.45? I am getting 0.82~.
• The calculation is sqrt((20 * π)/30) not just sqrt(20/30)
• 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 about , which he said only apply if acceleration is constant. Where does the last formula come from?
• 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.
• 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...
• 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!