Problem Solving Strategies

Suppose we call the bottom of the pulse height zero. That's the height at which the rope meets the pole if there is no pulse present. Then at time $t=0$ the height is zero. Yes?

Tanuj

yes

John Rennie

And at $t = 0.5$ the height is still zero. Yes?

Tanuj

yes

John Rennie

12:53 PM

Hmm, OK, I've just realised I get $v = 2$ cm/sec as well ...

Tanuj

yea

John Rennie

What is the (alleged) correct answer?

Tanuj

4cm/s

John Rennie

I don't understand why that is the correct answer.

Tanuj

I have an explanation with me too , (what they have provided), maybe you can judge by that whats the error they are making or we are making

John Rennie

12:57 PM

Yes?

Tanuj

particle speed = –$v_w$ × slope
= $2×(2/1)$ at
$\frac{3}{4}$ sec.

John Rennie

Huh? How can the slope be $2/1$? It goes up 1 cm in a horizontal distance of 1cm.

Tanuj

exactly !

lets put that on hold. I have one more question to ask.

Does it mean the disc is swinging left-right , like a pendulum seen from the front ?

John Rennie

I have to go I'm afraid.

Tanuj

Oh .. when will you be back ?

John Rennie

1:05 PM

The disk is swing left-right because it says it is oscillating in the plane of the disk.

Tanuj

just let me know how can I have a go at this and I'll try one more time

John Rennie

All you have to do is work out the restoring force at an angle $\theta$ then use the angular version of Newton's second law.

That will give you the equation of motion.

Tanuj

okay got it . When will you be back @JohnRennie ?

@blue do you have any clues about this one ?chat.stackexchange.com/transcript/message/43120180#43120180

001

1:24 PM

its 2 i think

Tanuj

@001 are you telling the answer to the first question ?

@skullpatrol Do you have any idea about this ?

skullpatrol

nope, sorry

Tanuj

its okay

Tanuj

2:23 PM

dudes , have a go at it

Gaurang Tandon

alright i'm trying this

1 hour ago, by Tanuj

@Tanuj Where do you get such interesting problems from?

i'm trying to figure out which component of tension to use, no luck

Tanuj

lol wouldn't it be $T\sin\theta$

Gaurang Tandon

i think the answer is C; the torque due to tension is zero about the pivot

am I right?

Tanuj

hell yea !

@GaurangTandon i just don't get what that means "the torque due to tension is zero about the pivot"

Gaurang Tandon

@Tanuj the pivot is the topmost point from where the string is suspended

Tanuj

2:34 PM

yes ! lol i know what pivot means

Gaurang Tandon

get it?

Tanuj

nope

how is the torque due to tension zero ?

Gaurang Tandon

why do you think its non-zero?

what point are you calculating the torque about?

Tanuj

about ahhh right

there are two !

idk man , I'm utterly confused

Gaurang Tandon

what point are you calculating the torque about?

answer this first ---^

Tanuj

2:38 PM

i thought it was just the one string , but yea there are two !

I don't know what axis of rotation do i use now

Gaurang Tandon

@Tanuj consider each point of suspension separately and then think

Tanuj

i was supposedly calculating it about that one suspension point

@GaurangTandon okay , I already did calculate the time period about one if you're asking that..

Gaurang Tandon

i was saying about the axis of rotation

Tanuj

I don't know @GaurangTandon I'm confused now

should it be one of the pivot points or the center of the disc or some other point ?

Gaurang Tandon

@Tanuj i really don't know how to explain my method

Tanuj

2:44 PM

hindi me

Gaurang Tandon

i don't know

Tanuj

lol what ?

just explain what do you mean by this- the torque due to tension is zero about the pivot

which point are you taking to be the pivot point ?

Gaurang Tandon

i don't know how to explain to you that I took each point as a pivot separately

Gaurang Tandon

3:11 PM

one way to think of it would be that since both strings are parallel and symmetrical wrt the disk's center, if one of them has a torque zero, then the other also has zero torque

the only torque is due to weight force only

but again, now I am myself confused between option A and C, A seems pretty valid as well

now I am stuck cries for help

2 hours later…

Abcd

5:16 PM

@Tanuj, @GaurangTandon Better just approximate the motion to be linear SHM, there were such problems in HCV.

So you get $2T \cos \theta = mg$

For very small $\theta$ , $\cos \theta = 1$

$\implies T = \dfrac{mg}{2}$

$F_x = 2T\sin\theta$

Suppose the pulley moved through an arc of length $x$

$\implies F_x = 2T \dfrac{x}{l}$

$ma =\dfrac{mg}{l}x $

$\implies \mathbb{Time \space{} period}= 2\pi \sqrt{\dfrac{l}{g}}$

Tanuj

5:30 PM

@Abcd okay got it

@Abcd lol you wouldn't believe so many genius minds got it wrong

and it was this straightforward

@Abcd why did you approximate it to be a linear SHM though ?

Abcd

@Tanuj This reminded me of another similar problem in HCV. The whole class wasn't getting the answer because everyone was trying it using angular shm.

Tanuj

It's quite crazy that you didn't even take the existence of disc into account

Abcd

@Tanuj For very small theta, the motion is linear, haven't you done such problems before?

Tanuj

I have , but that daunting disc did not let me accept that fact.

Hey man , I've not done HCV at all , just did DC Pandey .

@Abcd Thanks man :)

001

6:28 PM

this is not a one solutioon question

Abcd