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# Doubts ! Help Needed

Please give a solution to the below questions.

Note by Rajdeep Dhingra
4 months ago

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Congratulations for being selected to represent India in ijso. Also what about JOMPC? · 3 months, 4 weeks ago

Thanks. It will be out day after tomorrow. I have marked all the solutions I'll compile and post it on monday. · 3 months, 4 weeks ago

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.) . · 1 month ago

Cleared. Thanks anyways. P.S : your answer is correct. · 1 month ago

Hey @Rajdeep Dhingra why not coming to fiitjee ? PLease tell the truth !! · 1 month, 2 weeks ago

Which book is it ? · 3 months, 4 weeks ago

I don't know. Some old book for JEE subjective. · 3 months, 3 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. · 3 months, 4 weeks 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. · 3 months, 4 weeks ago

If he starts from June 10 , is he will be able to clear nsejs ? He also goes to coaching ... · 3 months, 3 weeks ago

I think so. · 3 months, 3 weeks ago

Solution for Question 1:

· 3 months, 4 weeks ago

Thanks ! · 3 months, 4 weeks ago

You are welcome · 3 months, 4 weeks 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 $$\dfrac{\omega}{2\pi} = \dfrac{1}{2\pi} \sqrt{\dfrac{2g}{3r}}$$ · 4 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 $$\dfrac{2g}{3r}$$. · 2 months, 3 weeks ago

Thanks · 3 months, 4 weeks ago