I think you mean the horizontal component and you would not take the horizontal component, because I think in the example above, we assume that the meter stick cannot move away from its set axis, and since the horizontal component means to go toward or away from the axis, it means the force is being canceled by the axis.
Addendum To really avoid confusion one should use R or RADIUS in your equations wherever you have used "m". Then its easy to see that the "m" following the Kg in the numerator is actually R*theta or R*radians. This allows theta or radians remain in the numerator. Otherwise you get only 1 in the numerator. After all angular acceleration is radians/s^2 and ang velocity = radians/s. Using "m" in place of R makes it confusing.
:) and there is no μs between the axis and the cylinder otherwise you would add a tangential force of friction and a normal force of friction with Ft ≤ μsFn if the rope had a mass then the mass of the cylinder would decrease over time you would then have mass not being a constant anymore thus derivating over time kinda like when you launch a rocket...without the jerk function ofcourse
Lever arm is 1 meter. Point of rotation is on the very left end, so the distance from the that point to the force, the weight of the object would be half of the whole length. Assuming the cylinder's center of mass is in the center.
hi monta vista ap physics 1 peeps :)))) hows the reading notes/vid lectures goin?
Thanks. My physics teacher doesn't teach and it makes the class 10x more work than it should be.
+Jacalyn Morgan ^^^FACTS
God Bless you Sir. Thanks for this service to humanity.
I love the sound of "br br br brbrbrbrbr"
if anyones in a hurry play at double speed. helps with cramming for exams!
@Haiwen Zhou TRUEEE
His sound effects are ridiculous. He's so awesome.
I think you mean the horizontal component and you would not take the horizontal component, because I think in the example above, we assume that the meter stick cannot move away from its set axis, and since the horizontal component means to go toward or away from the axis, it means the force is being canceled by the axis.
Addendum
To really avoid confusion one should use R or RADIUS in your equations wherever you have used "m". Then its easy to see that the "m" following the Kg in the numerator is actually R*theta or R*radians. This allows theta or radians remain in the numerator. Otherwise you get only 1 in the numerator. After all angular acceleration is radians/s^2 and ang velocity = radians/s. Using "m" in place of R makes it confusing.
@clespota Good point. We are assuming the string"s mass is negligible. That's physics lingo for the string's mass is zero. Very expensive string.
What is moment of inertia?
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The sound effects are really awesome. haha Thank you very much! Great video!
why does the pin not provide a force to the right at 7:04
I remember that the pin provided a force up AND right in a previous video
why not considering the vertical component 'mg sin theta' ??
:) and there is no μs between the axis and the cylinder otherwise you would add a tangential force of friction and a normal force of friction with Ft ≤ μsFn if the rope had a mass then the mass of the cylinder would decrease over time you would then have mass not being a constant anymore thus derivating over time kinda like when you launch a rocket...without the jerk function ofcourse
2:14 now i know how the rotation sounds like
how did u get L/2 for the value of R?
Lever arm is 1 meter. Point of rotation is on the very left end, so the distance from the that point to the force, the weight of the object would be half of the whole length. Assuming the cylinder's center of mass is in the center.
At 7:33 why is it L/2 and not just L?
The force causing torque and radius are already perpendicular. If they weren't then you would have to use the component that was perpendicular.
I recommend you to be a university lecturer. You are awesome!!
I'd be down to donate to this guy....he should have some sort of link to do that
I think my book says for a "thin rod axis perpendicular to rod and passing though center" as shown in video it is (1/12) not (1/3).
because mg acts at the center of mass of an object. He drew it at the center which would be half the length of the whole object. Therefore it is L/2
lol love the sound effects
Very nice explanation thank u so much sir.
i luwd the sound effects..:D awsumm video..!! thnkzz..
Great job sir thanks for this video
I think its because the rod is only able to apply force from its center of mass, which is where L/2.
excellent tutorial
very informative- thank u so much. just subscribed to you...
you rock buddy
Answer is 14.7rad/s^2
Jeez dude, making fun of someone for needing an explanation on a help video is like making fun of a sick guy for going to the doctor.
Thank you
Thank you!!!!
That's not what he was asking. He was asking about the force parallel on a ramp, which doesn't apply here.
ohh thats where light sabers get their sound from
lol
As someone who uses alpha for angles only, I cringed when you used that for torque
But it's used in physics formulas not as an angle but as alpha.
Its like physics ice.. Very expensive.. We need more grant money. For.. Physics ice..
The video keeps on blinking O.o
i.fucking.love.you
Ha ha sound effects!
haha!
abit slow but helpfull
mega noob detected
awful video