Angular Motion and Torque
ฝัง
- เผยแพร่เมื่อ 20 มี.ค. 2017
- More spinning things! Records, and wheels, and doors, and other fun things. The equations that govern this kind of motion are just like the ones we learned in kinematics, except they have different symbols. Lots of funky Greek letters in this one, folks.
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I like the way that you actually explain what torque is and not just how it is calculated
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i love how he gives us real life examples of usage of torque instead of just telling us that it's a thing that exists that we simply have to accept immediately
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"Clocks are negative" is how my textbook said to remember that clockwise motion is negative and hence counterclockwise is positive.
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Thak you!! Great video!! I was really happy to find info about inbetween perpendicular applied force and parallel applied force for torque
I had to skip this lesson to study for a calculus test I had the next day, and this video is extremely valuable to learn the topic, so good that I had to write a comment about it.
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Hi @ProfessorDaveExplains I have a few questions about torque that I don't know if you already covered in this video, because to be honest, I didn't watch it all, but you are the only one on youtube making videos on this topic that replies.
Here are my questions:
1. What absolutely decides whether rotation is clockwise and anti-clockwise? In a 3D spinning item, such as a bicycle wheel or even a ferris wheel like you said, has 2 sides. On one side, no matter how it spins, it will be clockwise, and the same wheel will appear to be anti-clockwise when viewed from the other side.
2. How does torque have a direction in or out of the page? I know the right hand rule RHR, but it doesn't explain anything. Taking the bicycle wheel or the ferris wheel as an example again, think about the axle: when looking at the wheel from the side so all you see is a thin rod, the axle sticks out on both the left and right, but when applying the RHR, and you pick a side, any side, your thumb will invariably end up pointing at either one of the left or the right... it's just a matter of which side you choose.
Further, even if the sides don't matter for points 1 and 2 because the mathematical signs will reverse which ultimately end up cancelling, what does it mean when people say torque has a direction in or out of the page? If a spoke in the bicycle rotates as the wheel spins, it spins either clock or anti clockwise depending on how you look at it. The force applied to it is perpendicular to the spoke on the same 2D plane as that of the flat surface of the wheel. However, if you used the right hand rule, your thumb will point in either direction of the axle that sicks out. But that's not where the force is, you are not pushing on the wheel from the side (remember we are looking at the wheel from the side so we only see the thin profile of the wheel tread and no spokes). What is there? How can you say that "torque comes out or into the page/plane" when torque is the force you applied to it (eg to the spoke) to make it turn?
Thank you in advance and sorry for the long question.
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Many people wonder why radians do not appear when we have radians * meters. Here is an attempt at an explanation:
Let s denote the length of an arc of a circle whose radius measures r.
If the arc subtends an angle measuring β = n°, we can pose a rule of three:
360° _______ 2 • 𝜋 • r
n° _______ s
Then
s = (n° / 360°) • 2 • 𝜋 • r
If β = 180° (which means that n = 180), then
s = (180° / 360°) • 2 • 𝜋 • r
The units "degrees" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r
s = 𝜋 • r
If the arc subtends an angle measuring β = θ rad, we can pose a rule of three:
2 • 𝜋 rad _______ 2 • 𝜋 • r
θ rad _______ s
Then
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
If β = 𝜋 rad (which means that θ = 𝜋), then
s = (𝜋 rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
The units "radians" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r
s = 𝜋 • r
If we take the formula with the angles measured in radians, we can simplify
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
s = θ • r
where θ denotes the number of radians (it does not have the unit "rad").
θ = β / (1 rad)
and θ is a dimensionless variable.
However, many consider θ to denote the measure of the angle and for the example believe that
θ = 𝜋 rad
and radians * meter results in meters.
Mathematics and Physics textbooks state that
s = θ • r
and then
θ = s / r
It seems that this formula leads to the error of believing that
1 rad = 1 m / m
and that the radian is a dimensionless derived unit as it appears in the International System of Units (SI).
In the formula
s = θ • r
the variable θ is a dimensionless variable, it is a number without units, it is the number of radians.
When confusing what θ represents in the formula, some mistakes are made in Physics in the units of certain quantities, such as angular speed.
My guess is that actually the angular speed ω is not measured in rad / s but in (rad / rad) / s = 1 / s = s^(-1).
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sir the unit of torque is NM that is joules. so work, energy and torque are the same quantities?
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This video is awesome and easy!!
I love your work so much, but if you can extend your videos to go a bit slow while explaining, thanks for your patience
try setting it to 0.75 speed if you need more time
there is a right hand rule for keeping track of positive or negative rotation that is much more robust and systematic than what he describes.
thank you so much!! : ))
Hey, would't the lever that will move the earth be faster than the speed of light since it will transfer force faster than light? So it will mean that something can travel faster than light?
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in your example the rod was horizontally placed , what if the rod was vertically placed ? then the force ( torque) will be perpendicular to the rod and consequently sine turns into cosine ??
It's always sine theta because theta is the angle between the force (torque) and the plane of rotating object (in other words , the displacement vector )and not the angle between the force with the horizontal. Torque is force times the perpendicular distance between the force and plane of rotating object.
Nice Tutorial, please how do I calculate this
A circular disc stacked has a total mass 2kg and rotates at the center (bearings).
Given that the diameter of the disc is 300mm, what amount of torque is required to make it rotate
continuously at 10 revolutions per second.
Trick question. No torque is necessary to make it rotate continuously at any rate, assuming ideal bearings support it. You need a torque to get it up to speed from rest, but no time interval for the initial acceleration was given.
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Hi - do I deduce correctly from your introduction (at 0:57) that uniform circular motion is translational motion? I quickly looked it up and one source says uniform circular motion is not translational motion, but rather a special form of rotational motion. I'm a bit confused with the new term "translational motion" in the context of your introduction. Can you please offer some explanation?
There are principles from both branches of classical mechanics, that apply to uniform circular motion. You can apply both the translational concepts to it, as well as the rotational concepts to it.
The term translational motion, just means that it is based on moving the object in its current form from position A to position B, without rotating it, extending it, or flipping it.
If an object were in pure translation during uniform circular motion, this would mean that one side of it constantly faces a fixed direction. So imagine painting a ball with half of it red, and the other half blue. Move it in a circular path, while a friend is watching you. Arrange it, so that the blue side always faces your stationary friend. This would be uniform circular motion in pure translation.
This usually doesn't happen, because it is a lot easier to create uniform circular motion in a way that it coexists with rotation at the same rate, such that one side of it constantly faces the center. It is a lot easier for you to rotate the ball, so that the blue side constantly faces you, rather than so the blue side constantly faces your stationary friend.
Wonderful
That transition at the end tho lol :)
For rotational kinematics, can any SUVAT equations be used for questions involving angular velocity, or only the 3 equations listed in the video?
The SUVAT equations all have counterpart equations in rotational form. They become θωαt equations (pronounced like throw-at, without the r), with the variables we traditionally use.
Here's the SUVAT equations for linear motion:
v = u + a*t
s = 1/2*(u+v)*t
s = 1/2*a*t^2 + u*t
v^2 = u^2 + 2*a*s
And here are their angular counterparts, respectively:
ω = ω0 + α*t
θ = 1/2*α*t^2 + ω0*t
θ = 1/2*(ωf + ω0)*t
ω^2 = ω0^2 + 2*α*θ
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2rsinc/t is it correct angular displacement c is angel
My guess is that actually the angular speed ω is not measured in rad / s but in (rad / rad) / s = 1 / s.
If we say that the measure β of the angle is θ radians, we mean β = θ rad, and θ is the number of radians (it does not have the unit "rad").
For emphasis we can say that θ is measured in rad / rad, since θ = β / (1 rad) and θ is a dimensionless variable.
When we use mathematical formulas to obtain formulas in Physics, we have to know what the variables represent.
What I consider a mistake, is present in the literature, it is not only in this video. Therefore, I will leave another comment with the detailed information to be examined.
When the video start I was sleepy, but as you arrive at Torque I stood up. KEA LEBOGA.
So if you have a disk of the same mass and you spin them so that they fly through the air, will be spinning disc fly further? Or less?
If you have two identical disks that you throw with the same linear speed, but give a spin to disk 1, and assume no air drag, they both will travel the same distance from the launch point to the first landing point.
Hey why didnt you incorporate the sin of the angle between the weight force and the seesaw in the end example, or is it insignificant
It's not insignificant. It is equal to 1 when the angle is 90 degrees, the largest value it can possibly equal. Due to this fact, multiplying by it has no impact on the equation.
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Sir, please make videos on moment of inertia
check out my tutorial on newton's first law of motion!
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in what cases angular momentum will not be in the direction of angular velocity....
can you make a video for that sir?!
Consider a rolling wheel of radius R, mass m, rotational inertia I about its center, and speed v of its center. The wheel is a uniform circular disk, which means I = 1/2*m*R^2. The wheel is rolling from right to left.
Its angular velocity is v/R.
Its linear momentum is m*v
We can pick any point we want, as the reference point to define its angular momentum. Pick a point that is in-line with the horizontal position of the wheel, but can be any height above or below the ground, which we will call Y. Positive Y is above the ground, and negative Y is below the ground.
The position of its center, relative to the reference point of angular momentum, is therefore at position R - Y. When Y = 0, the position of its center is R above our reference point. When Y is at the top of the wheel, Y=2*R and the position of its center is at -R. When Y=R, the reference point is at the wheel's center.
The angular momentum of its linear velocity is:
L_linear = (R - Y)*m*v
The rotational component of angular velocity is:
L_rotational = I*(v/R)
Put it together and get:
L = (R - Y)*m*v + I*v/R
L = (R - Y)*m*v + 1/2*m*v*R
L = m*v*(1/2*R - Y)
When we make Y = 1/2*R, the angular momentum about that point is zero. When Y is greater than this value, we have a negative angular momentum, which corresponds to clockwise rotation. But, our angular velocity of the wheel, is counterclockwise. The way we did this, was by relocating our reference point to define angular momentum, such that the linear component of angular momentum overtakes the rotational component.
In the last question why net torque must be zero?
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Thank you made me understand this,but sometimes are too fast when explain other topics like momentum
👍
I didn't understand why the direction of angular dispalcement is axial(along axis)?? Why do we right hand thumbrule??
The short answer is definition.
The long answer is that the linear kinematics quantities associated with rotation are difficult to track, and constantly changing direction. The one thing that remains constant, and is a reliable direction to define for angular kinematics and kinetics, is the axis of rotation. Therefore, we set up the mathematics around tracking these quantities as vectors about the axis, rather than the direction of any individual particle's motion.
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What's meant by fulkrum nd
How can planet orbits rotating without falling down is there any gravity involvement,Make my doubt cleared
Fulcrum means pivot point. It is the point where the lever is fixed in position, but free to rotate.
As for how a planet can orbit without falling down, the planet has sufficient tangential speed. It is continuously falling and overshooting the "target" that is causing it to fall. The force of gravity causes change in direction, rather than change in speed. There is gravity everywhere in the universe, so if someone tells you "there is no gravity in space", they are either oversimplifying something, or giving you incorrect information. There is gravity in space. But that gravity is "used up" in causing an object to orbit, which causes objects inside an orbiting vehicle to not need any constraint force. It can seem like they are in a zero gravity environment.
Consider what happens if you throw a ball perfectly horizontal. Initially, its velocity is horizontal. But gravity curves its path downward. Gravity is initially perpendicular to its velocity, where it can only cause a change in direction. Eventually, gravity starts to increase its speed, because it approaches the downward direction. What if you throw the ball so fast, that the Earth curves away from it, as fast as gravity can redirect its path? That is what orbit is. And if you neglect air drag, that speed would be 7.9 kilometers per second, and would circle the Earth in 84 minutes.
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thanks