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05:14
@StutiGupta Hi :-)
I'm only here until about 11 p.m. IST so you need to catch e before then.
Hi @JohnRennie
In Kavin's last question you said that the equilibrium extension is half the maximum extension. Will that hold for all cases, like a spring attached to a wall at one end, a spring-pulley system, etc? If not, how did you decide it was applicable in that case?
It applies to any system undergoing simple harmonic motion, and it's because the equilibrium point is exactly halfway between the two extremes.
Ah, I see it now.
Thanks
05:20
But you do have to think about the system e.g. with a spring attached to a wall the equilibrium point is when the spring is relaxed. Then the two extremes are an extension of A and a compression of -A.
So the equilibrium point is still halfway between the extremes, but unlike the mass hanging from a spring the equilibrium point is zero extension.
I'll just stick to equations then. Thanks.
OK :-)
@StutiGupta This method comes in useful when the equations are hard. That question about pulling the masses connected by a spring is a good example of where using the equations of motion is hard but using the equilibrium extension is easy.
05:52
@JohnRennie Oh, I see. Though if we were to give the first block an initial velocity and a force of 90N as well, then? (I'm sorry, I haven't done SHM yet.)
That makes it more complicated because then the initial state is not one of the extremes.
Alright, I'll try again after SHM. Thank you:)
@JohnRennie Hi !
Hi :-)
 
2 hours later…
07:33
@JohnRennie Sorry I got side tracked
@KavinIshwaran Hi :-)
Isn't this just τ = Iα ?
The instantaneous axis of rotation is the point of contact with the floor.
I get I to be 4MR^2
The MOI of the ring about its centre is I = 2mR²
07:42
2MR^2 ?
Yes
Then we use the parallel axis theorem to get I about the point of contact with the floor
So we add another 2mR² to get 4mR² in total
Then finally add another 2mR² to include the bead, making 6mR² in total.
Yes?
@JohnRennie I didn't get this
The bead is fixed to the ring, so when the ring rotates about the IAOR the bead rotates about the IAOR as well. Yes?
So the total MOI is I_ring + I_bead
Yes?
Are we agreed I_ring = 4mR²?
07:59
Yes
08:09
@JohnRennie I thought the for ring the IAOR will be about the point of contact and for the mass it will be about the center]
hence getting 5MR^2
I'm busy sorry. I won't be free for at least an hour.
Ok :-) please ping me when you are free
 
1 hour later…
09:23
@KavinIshwaran I for ring is about point of contact. You can visualise by imagining 3 scenarios
Imagine it's I is about centre then it would remain constant no matter where it is
So rolling ring with some acceleration would feel same everytime
But that's not true, If a person have to roll the ring on surface then it would take highest force in 1st case and lowest in 3rd and hence proved that we take MOI of mass m about point of contact.
09:39
@KavinIshwaran I'm back.
@KavinIshwaran No. everything is rotating about the IAOR so the distance of the bead from the IAOR is R√2. Hence Ibead = m(R√2)²
09:56
Ok then :-)
Does this make sense?
10:54
@JohnRennie Will you be around later today ?
@KavinIshwaran Yes, I'll be here around 10 p.m.
 
6 hours later…
17:05
@JohnRennie Hi !
@KavinIshwaran Hi :-)
Are you free ?
Yes :-)
Just one minute
What are the options?
Hello?
17:26
@JohnRennie Sorry for being slow in replying
whole day my network is troubling me
No problem :-)
@JohnRennie Just one minute
I sent the wrong image, that one is without option :-)
I am not sure with the C and D option
Any force that passes through the axis cannot change the angular momentum about that axis becaise the torque will be zero.
If the block loses contact with the cone the only forces acting are the spring and gravity.
The spring force acts through the axis and gravity acts downwards, parallel to the axis, so neither can change the angular momentum about the axis.
So I think C is true.
17:33
Yes it is given as the correct option
I'm not sure about D ...
Yeah it is given as correct also. I couldn't give a reason for that
r × v is the angular momentum (per unit mass) about the axis OO'.
Yes?
And the angular momentum about OO' is conserved because all the forces, spring, gravity, cone normal force, give a zero torque about OO'.
Yes?
17:37
Yes
So that means r × v must be conserved.
So D is true.
Does this make sense?
Yes :-)
I get it
OK :-)
@JohnRennie Thank you for the clarifications :-)
Will see you tomorrow :-)
Bye :-)

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