I though the vertical component of the tension in the string must be equal to the weight of the pendulum bob, because there is no acceleration in the vertical direction therefore Ft cos theta = mg I’m confused now please help me
so the tension should be constant right, but when I measured it in the lab using force sensor it save a sine graph on logger pro, what could be the potential cause for that.
the horizontal component is very confusing, although i can just blindly memorise it this way but why the horizontal force is not Rh=-Fc instead if the horizontal component of the weight is zero becuase if you resolve T then Tcostheta+Fc=mgcos90 so Tcostheta=-Fc? this gives me a headache...
at first, when the pendulum starts to spin, there is only tension in T1 but as you spin faster and faster the mass "wants" to rise up, but the lower string restricts the mass from doing so. when that happens tension starts to build up in the lower string.
no joke this might be the best video on youtube i found for explaining the conical pendulum, it was so easy and simple to understand
Thank you. You really touched on everything I needed.
Did I touch your heart though?
You touched EVERYTHING 😉
6:44 im not getting that value for T1.... i did exactly what u said
Yh I'm not getting it either
How to you find the radius if you don't know the speed?
h sin teta =r
Thank you so much... Really good explanation... I understand each and every point of the concept very well
How is T1 =67√3??? Supposed to be 59.47 N !! please check it🙏
yes
I though the vertical component of the tension in the string must be equal to the weight of the pendulum bob, because there is no acceleration in the vertical direction therefore Ft cos theta = mg I’m confused now please help me
Yes I'm confused too with that
hi can you fix the T1 value I think you got it wrong.
Thanks!
yes
thank you 😍😍
Thx❤️
Thanks ☺️
what if the string makes 90 degree with the vertical.... then which force balances mg??
That case is impossible :).
so that means.. can't we rotate the object horizontally to the ground???
we can not do so when the mass is under the influence of gravity
oh.. thanx
@@SimpleScienceProductions Even when the centripetal acceleration is high?
so the tension should be constant right, but when I measured it in the lab using force sensor it save a sine graph on logger pro, what could be the potential cause for that.
Bruce Xu interesting. Are you sure you were measuring the force of the tension? You may have been detecting acceleration or something else
Great video! Everything was clearly explained
W = mg
m = g/w?
@@simmyedits595 m= w/g
is it 5kg or 5g ?
Really helpful thank you so muchhhh
How do I solve questions like this with only the radius and alpha given, no given mass?
Late but the m's cancel!
He must have figured it out after 3 years lol
Thank you sir .. I really like your video
the horizontal component is very confusing, although i can just blindly memorise it this way but why the horizontal force is not Rh=-Fc instead if the horizontal component of the weight is zero becuase if you resolve T then Tcostheta+Fc=mgcos90
so
Tcostheta=-Fc?
this gives me a headache...
I want to ask .. what do we conclude from this?
i have homework and I have to write what I deduce
circ'lar and centripe'al and sim'lar??
You should explain T1 with 5g as it's 5*9.81(g)/sin(35)= 85.516N
how the tension is ocuuring in the string
at first, when the pendulum starts to spin, there is only tension in T1 but as you spin faster and faster the mass "wants" to rise up, but the lower string restricts the mass from doing so.
when that happens tension starts to build up in the lower string.
why are we working with kg no grams
You can convert it to grams simple
If we increase the mass of the pendulum it will not change the angle as one of the component of the tension will balance the weight right?
i'm wearing headphones and my ears got destroyed.
Some moments he's quiet and then all of a sudden he's screaming into the mic 😭😭
thank you so much for the wonderful explanation
awesome
Jiaosheng Cao