Congratulations for being selected to represent India in ijso.
Also what about JOMPC?
–
Shivam Jadhav
·
6 months, 1 week ago

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@Shivam Jadhav
–
Thanks. It will be out day after tomorrow. I have marked all the solutions I'll compile and post it on monday.
–
Rajdeep Dhingra
·
6 months, 1 week ago

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Bro is your second doubt cleared ? i am getting it's answer as root 5g/7(R-r) tell if not cleared i will give the solution ( if my answer is correct.) .
–
Shubham Dhull
·
3 months, 2 weeks ago

Hy rajdeep, can u give some tips for prepartion of nsejs , as my brother will be giving this year, presently he is in class 9.
–
Rahul Bubna
·
6 months, 1 week ago

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@Rahul Bubna
–
Read 9th and 10th class NCERT books thoroughly , if possible read 11th and 12th class NCERT also.(only relevant topics). For NSEJS most important is your speed of solving questions.
–
Rajdeep Dhingra
·
6 months, 1 week ago

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@Rajdeep Dhingra
–
If he starts from June 10 , is he will be able to clear nsejs ? He also goes to coaching ...
–
Rahul Bubna
·
6 months, 1 week ago

Question 2:
Let \( f \)denote the friction force acting on the body. Now, ablut the center of mass writing the torque equation,
\( f \cdot r = \dfrac{MR^{2}\alpha}{2} \)
\( f = \dfrac{MR\alpha}{2} \)
But since the cylinder doesn't slip \( r \alpha = a_{cm} \)
\( f = \dfrac{ma_{cm}}{2} \)
Now,
\( mg\sin(\theta) - f = ma_{cm} \)
\( mg\sin(\theta) = \dfrac{3ma_{cm}}{2} \)
\( a_{cm} = \dfrac{2g\sin(\theta)}{3} \)
\( \alpha = \dfrac{2g \sin(\theta)}{3r} \)
For small values of \( \theta \), \( \sin(\theta) \approx \theta \)
\( \alpha = - \dfrac{2g \theta}{3r} \)
The negative sign just denotes that direction of alpha is opposite to displacement.
This is the rotational analogue of \( a = -\omega^{2}x \)
\( \therefore \omega = \sqrt{\dfrac{2g}{3r}} \)
Frequency \( \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi} \sqrt{\dfrac{2g}{3r}} \)
–
Vighnesh Shenoy
·
6 months, 1 week ago

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@Vighnesh Shenoy
–
It is wrong bro. \( \theta\) is the angle you moved the center of mass not the angle by which you rotated the sphere. Hence you can't equate \( {\omega}^2 \) to \(\dfrac{2g}{3r}\).
–
Rajdeep Dhingra
·
5 months ago

## Comments

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TopNewestCongratulations for being selected to represent India in ijso. Also what about JOMPC? – Shivam Jadhav · 6 months, 1 week ago

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– Rajdeep Dhingra · 6 months, 1 week ago

Thanks. It will be out day after tomorrow. I have marked all the solutions I'll compile and post it on monday.Log in to reply

Bro is your second doubt cleared ? i am getting it's answer as root 5g/7(R-r) tell if not cleared i will give the solution ( if my answer is correct.) . – Shubham Dhull · 3 months, 2 weeks ago

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– Rajdeep Dhingra · 3 months, 2 weeks ago

Cleared. Thanks anyways. P.S : your answer is correct.Log in to reply

Hey @Rajdeep Dhingra why not coming to fiitjee ? PLease tell the truth !! – Parth Bhardwaj · 3 months, 3 weeks ago

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Which book is it ? – John Verghese · 6 months, 1 week ago

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– Rajdeep Dhingra · 6 months, 1 week ago

I don't know. Some old book for JEE subjective.Log in to reply

Hy rajdeep, can u give some tips for prepartion of nsejs , as my brother will be giving this year, presently he is in class 9. – Rahul Bubna · 6 months, 1 week ago

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– Rajdeep Dhingra · 6 months, 1 week ago

Read 9th and 10th class NCERT books thoroughly , if possible read 11th and 12th class NCERT also.(only relevant topics). For NSEJS most important is your speed of solving questions.Log in to reply

– Rahul Bubna · 6 months, 1 week ago

If he starts from June 10 , is he will be able to clear nsejs ? He also goes to coaching ...Log in to reply

– Rajdeep Dhingra · 6 months, 1 week ago

I think so.Log in to reply

Solution for Question 1:Log in to reply

– Rajdeep Dhingra · 6 months, 1 week ago

Thanks !Log in to reply

– Saarthak Marathe · 6 months, 1 week ago

You are welcomeLog in to reply

Question 2:

Let \( f \)denote the friction force acting on the body. Now, ablut the center of mass writing the torque equation,

\( f \cdot r = \dfrac{MR^{2}\alpha}{2} \)

\( f = \dfrac{MR\alpha}{2} \)

But since the cylinder doesn't slip \( r \alpha = a_{cm} \)

\( f = \dfrac{ma_{cm}}{2} \) Now,

\( mg\sin(\theta) - f = ma_{cm} \)

\( mg\sin(\theta) = \dfrac{3ma_{cm}}{2} \)

\( a_{cm} = \dfrac{2g\sin(\theta)}{3} \)

\( \alpha = \dfrac{2g \sin(\theta)}{3r} \)

For small values of \( \theta \), \( \sin(\theta) \approx \theta \)

\( \alpha = - \dfrac{2g \theta}{3r} \)

The negative sign just denotes that direction of alpha is opposite to displacement.

This is the rotational analogue of \( a = -\omega^{2}x \)

\( \therefore \omega = \sqrt{\dfrac{2g}{3r}} \)

Frequency \( \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi} \sqrt{\dfrac{2g}{3r}} \) – Vighnesh Shenoy · 6 months, 1 week ago

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– Rajdeep Dhingra · 5 months ago

It is wrong bro. \( \theta\) is the angle you moved the center of mass not the angle by which you rotated the sphere. Hence you can't equate \( {\omega}^2 \) to \(\dfrac{2g}{3r}\).Log in to reply

– Shubham Dhull · 1 week, 4 days ago

it would be 5g/7r instead !Log in to reply

– Rajdeep Dhingra · 6 months, 1 week ago

ThanksLog in to reply

@aryan goyat@Prakhar Bindal@Archit Agrawal@Aniket Sanghi Please help – Rajdeep Dhingra · 6 months, 1 week ago

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