IIT JEE , AIEEE Forum - Physics , Mechanics

doubts.com1

thank you very much............goiitians.............at last i didnot get the answer.

doubts.com1

thanx goiitians............at last i didnot get the answer.......................

varun.tinkle

JUST WANTED T KNOW WHATS WRONG IN MY WORKING

shreyasnivas

nope. it rolls instantly..

how can you say such a thing man?

we are talking about an ideal situation.. not something in real life. dont waste the community space with silly pointless arguments...

varun.tinkle

ACCORDING TO THE QUESTION IN THE STARTING IT DOES NOT ROL
WHAT DO U SAY ABT THAT

shreyasnivas

look man.. its cool if you are really tring to help whoever is asking the doubts.. but dont argue when you are wrong.. you know well that as soon as the impulse is given, it will TRY to translaet, but the friction will INSTANTLY make it roll and slip.

now i will ask you politely to stop saying nonsense. dont think that whatever you say is right. i dont blame you for trying to prove yourself, but please get your concepts clear before you argue.

goiit is not made for people who want to act cool and stylish.. its to help students. i really congratulate you if you are in the 8th standard.. but if you are not.. then you are fooling only yourself.

if you really are in the 8th standard , i respect you for trying.. if you can do so much at this age.. then you can be AIR 1.

varun.tinkle

DUDE I KNOW WHAT R U TRYING TO TELL
BUT MY DOUBT IS THAT WONT THE IMPULSE ALSO HV A TURNING EFFECT SO THE SPHERE WILL TRY TO SLIP BACKWARDS THEN FRICTION WILL ACT IN THE FORWARD DIRECTION
AND ACCORDING TO ME BOTH THE MAG OF THE TORQUES ARE EQUAL SO THE ODY DOES NOT ROTATE IN THE STARTING

shreyasnivas

since the duration of the impulse is not given (it generally is not given in such problems) we assume that the impulse is for almost zero time.. otherwise it is impossible to solve this question.

varun.tinkle

SO IN THAT SITUATION THE SPHERE WILLL ROTATE ALMOST INSTANTLY
AM I RIGHT

elessar_iitkgp

The length of the planck, L = 12 m

Initial velocity of the cylinder, ${v}_{0}$ = 7 m/s

Coefficient of friction between the cylinder and plank, $\mu$ = 0.1

Friction acts on the cylinder in a backwards direction in its full magnitude.

Equations:

$m{a}_{1} = \mu mg$

$\frac{1}{2} m{R}^{2} \alpha=\mu mgR$

$2m{a}_{2}=\mu mg$

Let at anytime t, the velocity of cylinder be ${v}_{1}$ , its angualr velocity be $\omega$ and the velocity of plank be ${v}_{2}$ . Let at time T the slipping is stopped. Then,

${v}_{1}(T)-R\omega(T)={v}_{2}(T)$

${v}_{0}-\mu gT-2\mu gT = (\mu g/2)T$

$T=(2{v}_{0})/(7\mu g)$

Substituting the values,

In these two seconds the cylinder has travelled the following distance on the plank

${L}_{1} = {v}_{0}T-(1/2){a}_{1,2}{T}^{2}$

where ${a}_{1,2}$ is the accelaration of the cylinder wrt the plank

Substituting values, ${L}_{1} = 11 m$

After the time , friction stops acting and the cylinder and the plank move with constant velocities

${v}_{1}={v}_{0}-\mu gT=5 m/s$

and

${v}_{2}=(\mu g/2)T = 1 m/s$

So time taken by the cylinder to cover the additional remaining is

$T'=(L-{L}_{1})/{v}_{1,2} = (1/4) s$

${v}_{1,2}$ being the relative velocity of the cylinder wrt the plank

So total time taken

varun.tinkle

ELESSAR
THE STARTING STEPS OF UR WORKING FOR SAME FOR ME
BUT I CANT UNDERSTAND HOW
v1(T)-RW(T)=v2(T)

varun.tinkle

OK NOW I UNDERSTOOD

doubts.com1

thank you elessar..............thank u very much....................

as i am new to goiit i dont know how to rate you.

Ask & Discuss Questions with Community & Experts