10-05-10: Mechanics

Sibal I have the rights to allow people to post questions on QOD..

THere are a lot of users here (generally old ones who have this right... )

See the past.. lots of users including Prophet sir, Kaymant sir, Abhishek priyam, Celestine and eureka, Bhargav have posted these....

ok if not the full soln then let me start the initial analysing...tell me wethr i am right or wrong...

frstly due to the inertia the cylnder will try to remain in its position whilst the block will move past...due to this sliding the friction comes into play and the ball starts rotating in anticlockwise orientation...[1]

now we have to consider the translational part...

supposing the inertia thing is correct consideration we move forward like this...

let velocity of the block initially is v....

the friction acting on it in backward direction...let friction=f

so acceleration in backward direction= f/2M

now if the cylinder remains in its own position whilst the block translates then time taken for the cylinder to fall off is time taken by the block to move forward by d distance

now, d=vt-¹/₂at²..

this quadratic gives us the time t...

now i have a doubt wether or not the consideraton that friction f remains equal throughout the journey period is true or not...[3]

next for the cylinder, angular acceleration α=Γ/I=frMr²=fMr=2a/r

so in the time t...found out in previous to previous post...

ω=αt=ftMr=2at/r

pure rolling...

V_final=rω_final

→ v-ft2M=rftMr=ftM

→v=3ft2M

now the most important part....how to find out the value or v...

that is after the collision occurs..

so for totally elastic collisions, e=1

and ΔEnergy=0

so, velocity of separationvelocity of approach=1

or, v+v'=v_o [+ since the mass gets reflected back...]

and 12Mv_o²=12.2Mv²+12Mv'²

so that gives v

and ω is ignored in energy expression becoz ω=0 at time the small mass hits the block...

so except for the numerical part i thnk theoritically the problem is done...[1][1][1]...yay!!

solving, v=-23v_o

sorry i considered solid cylinder in this problem instead of hollow one...

so wrote I=12Mr²...

so i ll have to edit evrythng...will do it in few mins....please bear till then....

*edited...now i thnk it is ok...

maximum v_o=9ft4M

again maximum case achieved when friction is limiting...i.e μMg for this case...

thus

max v_o=94μgt

now i have a problem...i am getting two vlues of t...since it is obtained from a quadratic...

now to maximize v_o the value with (+) will give max but when the time wid (-) is achieved the cylinder falls off.....so i thnk we shud take that value only!!!

t=2v±√4v²-8ad2a=v±√v²-2ada...[1][1]

and apologies......the actual soln is not so FAT as it eems...it is slim...but as at that time i din have pen and paper so i had posted every single thing that came to mind here!!!!

points to be debated...

1) does the cylinder remain in the starting position throughout the journey till it falls off?

2) does friction remain unaltered throughout the journey?

3) which time to be used? the one wid (+) or the one wid (-)?

i hope that so far the forum was not active in this thread but now the forum will get active in this thread!!...(i really hope so!!)