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1. A pendulum bob of mass 1.0 kg is attached to a string 1.0m long and made to revolve in a horizontal cycle of radius 0.6m. Find the period of the motion and the tension of -2 the string (assume g = 10 ms ).
What if I don't have a time and the only thing I got is the start speed, end speed, the chord and the radius? Can I know the tangential acceleration then?
how do I taugle this... 1. A pendulum bob of mass 1.0 kg is attached to a string 1.0m long and made to revolve in a horizontal cycle of radius 0.6m. Find the period of the motion and the tension of -2 the string (assume g = 10 ms ).
I think it’s important to add, that with the first example, the net acceleration calculated is in respect to 5 seconds with the speed of 40m/s. Though tangental Acceleration would be constant, the Fc increases with time, so if he calculated the Fc at a speed different to 40, let’s say 20m/s, we would get a lower Fc, so the net acceleration would be the sq rt. of 8^2 [our constant Tangental Ac] . 20^2/800[ Our Contripetal Ac at the speed 20m/s.
Tangential speed (omega) = V(perpendicular)/radius Tangential acc (alpha)= a(tangential)/radius The difference between tangential acc and tangential speed is that the tangential acc is same acceleration that you study in kinematics and the tangential velocity is not dependent on tangential acc although tangential acceleration can change tangential speed . The relationship is same as studied in kinematics that the object can have constant velocity even in absence of acceleration. Only fact that should be noted both in kinematics and circular motion that velocity is constant( only magnitude in case of circular motion) when tangential acceleration is not present. Hope you got what you wanted . If not reply asap 🙏😊🙏
Net force matches the direction of the centripetal acceleration. No way to calculate angle because there is no tangential acceleration because the particular moves at constant speed. The net force component is simply the acceleration component times the mass I think
Can somebody please explain why the centripetal acceleration formula from uniform motion also applies to non uniform motion? The derivation I see everywhere comes from the assumption that |v1| = |v2|. Derivation in uniform motion: IF 1) |v1| = |v2| = |v| and v2-v1 = delta_v in common base 2) where delta_s = the vector between the bases of v1 and v2 THEN delta_s / R = delta_v / v rearranging: delta_v = v * delta_s / R dividing by time: a = delta_v / delta_t = v * (delta_s / delta_t) / R = v^2 / R How is it proven that the same geometrical constraint is true in regard to centripetal acceleration if it's not true that |v1| = |v2|?
very interesting and eazy to understand ; i think the like is less because the time your'e taking to draw those things, try to use any other softwares which looks better than this.
Bro . When the tangential acceleration is present speed and velocity(i mean direction and magnitude both) are not constant . That is the reason why centripetal acceleration is also not constant . And also the net acceleration is= root (a tangential ^2 + a centripetal. ^2 ) So this means that the object will still have a circular motion . BASICALLY YOU DON'T NEED TO WORRY ABOUT THAT . THE ABOVE EQUATION OF NET ACCELERATION WILL ADJUST ON ITS OWN
Hey Organic Chemistry Tutor. I wanna marry you. I love you so much on god. I been trying to use a textbook to help me with this concept for an hour and I just watched 7 minutes of this video and understand way more than I did with the book. basically all i am saying is I wanna have ur kids *muah*
So on a uniform circular path, you can actually write the position of a particular with a vector r having the coordinates r as the length of the radius and and theta as the positive angle with the X axis. At time t=1 and t=2 the object will be at two different locations and the difference is the vector dr. The same is also true for their velocity vectors just as the position vector. Since the speed, i.e. the length, the magnitude of the velocities v1 and V2 is same and the difference between the two vectors is dv. The angle is also theta. As the angles are the same theta and both triangles have two sides of equal length, (The first triangle has the side length r, and the second triangle v), the triangles are similar and one can write v/dv=r/dr so dv=v/r*dr. And recall the acceleration can also be written as dv/dt as the change in velocity divided by the change in time. If you put v/r*dr in the place of dv in the second equation you get (v/r)*(dr/dt). Recall how we defined r as a position vector. So dr/dt is the change in position divided by change in time which is nothing other than velocity. So the equation becomes a=v/r*v = v^2/r
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Whenever i search up a concept im struggling in, I always hope your name pops up to teach me
Same
Same
same
Same
aren't we all😅
Thank you. Not a lot of videos about non uniform that actually goes into the math much
You are the best
I got A+ in Physics because of you .
Professor did not explain it like this! super helpful
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You are the best
What if the object didn't start from rest
which speed do i use to calculate the centripetal acceleration?
great great great great vedio and helps a lot on my test!
The best lecture ever✨
Indeed
You r so helpful ❤️
1. A pendulum bob of mass 1.0 kg is attached to a string 1.0m long and made to revolve
in a horizontal cycle of radius 0.6m. Find the period of the motion and the tension of
-2
the string (assume g = 10 ms
).
Period is equal to 1 divided by the frequency (rotations/min). Hope this helps
another great video. thanks for all the help
Would be nice if you had a video on the Corliolis-acceleration too ;-)
V good ❤❤❤
Nice explanation
Good luck sa mga nag exam dira 😂👌🔥🇵🇭
hahahahahaha
daghan salamat unta tanan
I’m confused. I thought tangential acceleration was r*alpha? Why did he put it as (V_F - V_0)/t?
What if I don't have a time and the only thing I got is the start speed, end speed, the chord and the radius? Can I know the tangential acceleration then?
yes.
2as = vf² - vi²
a = what u wanna know
s = chord = angular displacement
how do I taugle this...
1. A pendulum bob of mass 1.0 kg is attached to a string 1.0m long and made to revolve
in a horizontal cycle of radius 0.6m. Find the period of the motion and the tension of
-2
the string (assume g = 10 ms
).
11:27 why is angle in degrees and not in radians?
YAAAANNNN STEM PAAAAAA
Thanks
fing thank u man!
Sir ac=acostheta please explain what the value of theta will be?
I wish you could replace my prof.
fire my professor and hire this guy
Thank you for telling this matter.
Thank you as your service.
gamed ya basha
What does the angle theta represent?, the angle with respect to what?
I think it’s important to add, that with the first example, the net acceleration calculated is in respect to 5 seconds with the speed of 40m/s. Though tangental Acceleration would be constant, the Fc increases with time, so if he calculated the Fc at a speed different to 40, let’s say 20m/s, we would get a lower Fc, so the net acceleration would be the sq rt. of 8^2 [our constant Tangental Ac] . 20^2/800[ Our Contripetal Ac at the speed 20m/s.
The object is moving with a constant rate of change in velocity but the rate of change in speed is variable. Is this type of motion possible ??
What’s the difference between tangential speed and acceleration and what’s the formulae or how do u find tangential speed if there is a difference?
Tangential speed (omega) = V(perpendicular)/radius
Tangential acc (alpha)= a(tangential)/radius
The difference between tangential acc and tangential speed is that the tangential acc is same acceleration that you study in kinematics and the tangential velocity is not dependent on tangential acc although tangential acceleration can change tangential speed . The relationship is same as studied in kinematics that the object can have constant velocity even in absence of acceleration. Only fact that should be noted both in kinematics and circular motion that velocity is constant( only magnitude in case of circular motion) when tangential acceleration is not present. Hope you got what you wanted . If not reply asap 🙏😊🙏
Done
What is the direction of net force in non uniform accerlation? Can we calculate the angle in this case?
Net force matches the direction of the centripetal acceleration. No way to calculate angle because there is no tangential acceleration because the particular moves at constant speed. The net force component is simply the acceleration component times the mass I think
Can somebody please explain why the centripetal acceleration formula from uniform motion also applies to non uniform motion? The derivation I see everywhere comes from the assumption that |v1| = |v2|.
Derivation in uniform motion:
IF
1) |v1| = |v2| = |v| and v2-v1 = delta_v in common base
2) where delta_s = the vector between the bases of v1 and v2
THEN
delta_s / R = delta_v / v
rearranging: delta_v = v * delta_s / R
dividing by time: a = delta_v / delta_t = v * (delta_s / delta_t) / R = v^2 / R
How is it proven that the same geometrical constraint is true in regard to centripetal acceleration if it's not true that |v1| = |v2|?
I have this problem. Now of you know about that please explain it to me.
very interesting and eazy to understand ; i think the like is less because the time your'e taking to draw those things, try to use any other softwares which looks better than this.
That's not true. This is better 😐
My goat
so much better than the stupid textbook
Organic Master
whats net acceleration for? it's not in my school material...
the angle seems wrong, well I guess it also depends on point of refernece? am I wrong?
When there is a tangential acceleration, the velocity is changing. Then how can the centrepetal acceleration become constant (v2/r)..?
Sreenadh E K that's a good question 🤔
I guess the v there is the velocity at a certain point in time
Bro . When the tangential acceleration is present speed and velocity(i mean direction and magnitude both) are not constant . That is the reason why centripetal acceleration is also not constant . And also the net acceleration is= root (a tangential ^2 + a centripetal. ^2 )
So this means that the object will still have a circular motion . BASICALLY YOU DON'T NEED TO WORRY ABOUT THAT . THE ABOVE EQUATION OF NET ACCELERATION WILL ADJUST ON ITS OWN
Hey Organic Chemistry Tutor. I wanna marry you. I love you so much on god. I been trying to use a textbook to help me with this concept for an hour and I just watched 7 minutes of this video and understand way more than I did with the book. basically all i am saying is I wanna have ur kids *muah*
u didnt point out that the radial acceleration is the opposite of centripeteral acceleration
where does v²/r came from?
It's a law. Memorize it!
So on a uniform circular path, you can actually write the position of a particular with a vector r having the coordinates r as the length of the radius and and theta as the positive angle with the X axis. At time t=1 and t=2 the object will be at two different locations and the difference is the vector dr. The same is also true for their velocity vectors just as the position vector. Since the speed, i.e. the length, the magnitude of the velocities v1 and V2 is same and the difference between the two vectors is dv. The angle is also theta. As the angles are the same theta and both triangles have two sides of equal length, (The first triangle has the side length r, and the second triangle v), the triangles are similar and one can write v/dv=r/dr so dv=v/r*dr. And recall the acceleration can also be written as dv/dt as the change in velocity divided by the change in time. If you put v/r*dr in the place of dv in the second equation you get (v/r)*(dr/dt). Recall how we defined r as a position vector. So dr/dt is the change in position divided by change in time which is nothing other than velocity. So the equation becomes a=v/r*v = v^2/r
Dropping out…
can I replace v²/r by At = Ac×r ?
ok but what if the problem is in radians?
Are you saying "tan-genital"?