Yesterday Philip was frustrated at the lack of gud problems...I am posting some for everyones practice..........2009 aspirants should solve them and put up solutions here for 2010 students to learn.............
Some might be easy.......but all are conceptual[1][1].....
Q1 A uniform rod of length 2a and mass m is rotating in horz plane about a smooth fixed pivot through the centre of rod with a constant angular velocity w rad/s.A particle of same mass moving with speed aw/4 opposite t orod strikes one end of the rod perpendicular to it.If e=1/2,calculate angular velocity of rod immediately after collision.
If anyone has problem with my initiative he/she is free to express his/her views...........i wont feel bad at all and i will stop this work at that moment only..[1][1][1][1]
24Q5 2 cylinders having radius R1 and R2(R2>R1) and rotational inertia I1 and I2 respectively are supported by fixed axis perpendicular to the plane.The large cylinder is initially rotating with angular velocity w1.The small cylinder is moved to the right until it touches the larger cylinder and is caused to rotate at constant rates in opp directions.Find angular velocity w2 of small cylinder.
1equal and opposite impulse should do it i think
*** edited *** impulse is equal but moment of impulse isn't so the rest of the soln hidden is wrong as pointed out by eureka
I1 w2 = I2 (w1-w2) (assuming of course that friction is there)
like this
is this correct ??
so maybe w2 = I2 w1
I1+I2
let equal impulse be J
then JR2 = I2 (w1 -w') w2 ( hehe wrong bcoz w' ≠w2)
and JR1 = I1 w2
now over here w'R2 = w2R1
see iitimcoming's complete solution below....
he deserves the credit..
1atleast check the answer i'll hide it if its correct
now it is correct [1]
24sorry......in the answer R1 and R2 terms are also there.................ok got to go......bbye......[13]
1which torque sankara the system comtains both bodies and the sys is horizontal so which torque ?
3appling the conservation of ang.momentum abt the center of axis fo the larger one.
I1w1=I1w'+I2w2
and also R1w2=R2w' since there is no slipping at the pt of contact.
w'=R1w2/R2
substituting the w' in the first eqn gives the value of w2.
eureka pls check dis method.
3SANKARA HOPE U REALISED THAT THERES TORQUE DUE TO FRICTION ABT THE POINT UVE CHOSEN........[1]
24bingo........i was waiting for this only.....[1][1].....good job..iit....[1][1][1]
1i dont know iitimcoming but your method also involves impulse......
anyway my soln has a mistake and i admit it I'll edit that small mistake now.. thanks for telling though...[1]
u are right abt the variables given in the ques they're quite ulta
3if its along the plane slantin from left to right..............
then normal component of the force is Fcosθ...........
area of crossection is A/cosθ!!!!!!!!!!!
Fcos2θ/A...............
SHEARING STRESS=
Fcosθsinθ/A................
if that plane is variable then we can find in which plane the stress is max........
Fsin2θ/2A...........
max value is when 2θ = pi/2............
θ = pi/4!!!!!!!!!!!!!!!!!!!!
and for tensile stress its max when θ=0.......
24here is Q7
A satellite of mass 2m is in a circular obit of radius 2R around earth.By mistake another satellite of mass m is put into the same orbit. having opp sense of rotation.The collision b/w 2 satellites is perfectly inelastic. Will the combined masss hit the earth surface?If,yes at what angle to the horz?If no,what is its min distance from surface of earth
24in Q6 Aeff=A/cosθ was the most imp thing........rest was easy.....
3sry for posting my question in this thread.
for the last qn
u=v/3
and then it wil be like a projectile having initial tangential velocity.
3yes itll be at v/3 velocityy.............= 1/3√GM/2R
the velocity alnog other dir when it reaches earth ......
GM/r2 = vdv/dx ...............
integrating with appropriate limits we get v = √GM/R .............. so
θ = tan-1[3√2] .........
24Q8 ka mere paas answer nahin hai.......par theek hi lag raha hai..........
Q7 ka solution galat hai.............recheck it....
24Well everyone would be thinking that this thread is closed.............but no .....its still alive..........i was busy with chem boards.....which went horribly bad[2][2].....hope it doesnt affect jee
Q9 
