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Abcd
Abcd
1:06 AM
@AvnishKabaj ^
 
Avnish Kabaj
Avnish Kabaj
1:47 AM
@Abcd is $g - \frac{\alpha(l/6 -l\cdot sin^2\theta)}{sin\theta}$
The answer
 
Abcd
Abcd
@AvnishKabaj Nope :/
 
Avnish Kabaj
Avnish Kabaj
What's the answer
@Abcd ansaaaa
 
Abcd
Abcd
@AvnishKabaj $a= \dfrac l 2 \sin\theta \alpha + \dfrac l 2 \omega^2 \cos \theta $
 
 
1 hour later…
Avnish Kabaj
Avnish Kabaj
3:03 AM
@Abcd are you sure the second term is there
 
Abcd
Abcd
3:21 AM
@AvnishKabaj yes
 
Abcd
Abcd
3:43 AM
Does anyone remember/ know that formula that relates coefficient of restitution's square with change in kinetic energy during collision? @AvnishKabaj @GaurangTandon @Tanuj
 
Gaurang Tandon
Gaurang Tandon
4:04 AM
@Abcd it wasn't supposed to be rote learned iirc; i'd just derive it
 
Abcd
Abcd
@GaurangTandon are we referring to the same formula?
@GaurangTandon It reduces 4-6 steps!!
 
Gaurang Tandon
Gaurang Tandon
i know, well i didn't rote learn it
 
Abcd
Abcd
I see
 
 
1 hour later…
Abcd
Abcd
5:27 AM
Hey @JohnRennie !!
 
 
1 hour later…
Abcd
Abcd
6:41 AM
@JohnRennie Are you there?
 
John Rennie
John Rennie
I'm pretty busy this morning and won't be available for a while. Sorry :-(
 
Abcd
Abcd
Oh, fine.
 
Abcd
Abcd
7:04 AM
 
Nehal Samee
Nehal Samee
7:55 AM
For not skidding , ...I think...
@Abcd...Wcosθ=μN+F_{c}sinθ...
F_{c} is centrifugal force ...
For the other one...
Wcosθ+μN=F_{c} sinθ...
 
 
1 hour later…
Abcd
Abcd
9:00 AM
Let me retry here:
Friction = $f$
@NehalSamee You have written the same equation twice.
$f- mg\cos \theta + m\omega^2 \sin \theta = 0$
Agh! no
$f - mg\cos \theta + mr\omega^2 \sin \theta = 0 $
Then:
$N - mr\omega^2 \cos \theta - mg \sin\theta = 0 $
$f_{max}= \mu N $
$\mu({-mr\omega^2 \cos \theta + mg \sin\theta})= mg\cos \theta - mr\omega^2 \sin \theta $
 
Nehal Samee
Nehal Samee
9:18 AM
Hope it helped @Abcd...
 
Abcd
Abcd
Oh my last equation is wrong,
Correction:
$\mu ({mg \sin \theta+ m r \omega^2 \cos \theta })= mg \cos \theta - mr\omega^2 \sin \theta $
On solving,
$\omega = \sqrt{\dfrac{g- \mu g \tan \theta}{\mu r + r\tan \theta}}$
$f_{min}= 0$
Voila! $\omega_{min}= 0.204$
Voila! $\omega_{max}= 0.5 $
Done.
@NehalSamee Ty for your contribution too.
 
 
3 hours later…
Gaurang Tandon
Gaurang Tandon
12:47 PM
just post
 
Abcd
Abcd
in The h Bar, 2 mins ago, by Abcd
@JohnRennie The moment of inertia of a square of side a about its diagonal is $\dfrac{Ma^2}{12}$. Now if I cut it through the diagonal into two equal triangular pieces, wouldn't the moment of inertia of each about that diagonal be $\dfrac{Ma^2}{24}$? (coz MI is a scalar quantity)
 
Gaurang Tandon
Gaurang Tandon
i'm tempted to say Yes
 
Abcd
Abcd
@GaurangTandon But see this question:
Never mind, I did it.
 
Gaurang Tandon
Gaurang Tandon
so what was the answer?
 
Abcd
Abcd
@GaurangTandon A
 
Gaurang Tandon
Gaurang Tandon
12:52 PM
plz explain how
 
Abcd
Abcd
It would be very tough with integration.
@GaurangTandon See consider a full square plate of mass 2M
 
Gaurang Tandon
Gaurang Tandon
oh ok got it
yes yes got it
 
Abcd
Abcd
Yeah, thats the silly mistake I was making !!
54 secs ago, by Abcd
It would be very tough with integration.
@GaurangTandon can you do this with integration?
 
Gaurang Tandon
Gaurang Tandon
i mean I can, but I don't want to
super boring
 
Abcd
Abcd
I see.
 
 
3 hours later…
Abcd
Abcd
4:13 PM
@JohnRennie Quick question...
 
John Rennie
John Rennie
@Abcd Yes?
 
Abcd
Abcd
@JohnRennie I am trying to use:
$(v_o- at)= R(\omega_o+ \alpha t )$
But why doesn't it give the right answer,
I get $7v_o/9$
 
John Rennie
John Rennie
I'm not sure offhand how to approach that. It seems simple enough ...
 
Abcd
Abcd
2 mins ago, by Abcd
$(v_o- at)= R(\omega_o+ \alpha t )$
Does this make sense?
$v_o - \dfrac f m t = R (v_o/(3r) + I/(fR)t $
Where f is friction.
I'll get value of $t$
Then I'll put it in velocity
 
John Rennie
John Rennie
I see what you're saying. The frictional force provides both a linear impulse and an angular impulse.
 
Abcd
Abcd
4:22 PM
Yes
And for pure rolling $v= R\omega$
 
John Rennie
John Rennie
Force times time is the change of linear momentum and torque time time is the change of angular momentum. That's probably the way I'd approach it.
 
Abcd
Abcd
Why is my method wrong?
$f= \mu m g$
On solving $\dfrac {2v_o}{9\mu g}= t$
$v= v_o - (\mu m g/m )\dfrac{2v_o}{9\mu g}$
 
John Rennie
John Rennie
What's I for a disk again?
 
Abcd
Abcd
$\dfrac {1}{2} mR^2$
$\implies v= \dfrac {7v_o}9$
This is bad :"( . as well as sad. How can physics yield different answers with different methods :/
 
John Rennie
John Rennie
Hmm, I just got $\tfrac{7}{9}v_0$ as well ...
 
Abcd
Abcd
4:36 PM
let me show you how they have done it:
@JohnRennie ^^^
@JohnRennie Is this their mistake^^^ ?
Have they taken the wrong signs?
 
John Rennie
John Rennie
If we rewrite their equation we get: $$ mrv_f - mrv_i = -I\omega_f - I\omega_0 $$
That doesn't seem right ...
 
Abcd
Abcd
Okay...
 

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