IIT JEE , AIEEE Forum - Physics , Mechanics

paulparthapratim2

I cut pasted it frm somewhere else.

Think m as a 2 kg block and where is 2 kg in the picture , assume it to be 4 kg.

If the friction coeff b/w 2kg block ( in pic m ) and plane below it is 0.5 and the plane below 4 kg block ( in pic which is 2 kg ) is frictionless,.

pulley has radius 10 cm and MI 0.5 kg-msq.

what is the accln of 4 kg block ?

varun.tinkle

CANT SEE THE PIC

nunknown91

yes cant see the pic

varun.tinkle

DONT CUT AND PASTE
ATTATCH THE FILE

sriram.a

$\frac{4g}{\sqrt{2}}-{T}_{1}=4a$

${T}_{2}-(\frac{2g}{\sqrt{2}}+\frac{g}{\sqrt{2}})$

-------------------------------------

${T}_{2}-{T}_{1}=6a-\frac{g}{\sqrt{2}}$

$({T}_{1}-{T}_{2})0.1=0.5\alpha$

now solve above equations

varun.tinkle

SEE
MGSIN45-T=MA (FOR 4 KG BLOCK)
40/ROOT 2-T=4A

T-MGSIN 45 -UMGCOS45=MA (FOR 2KG BLOCK)
T-20/ROOT2-10/ROOT=2A
T-30/ROOT2=2A

T1R-T2R=I A(ANGULAR ACC)
T1R-T2R=1A/R
T1-T2=IA/R^2
TI-T2=50 (COMPUTING THE VALUES)

T1-T2=IA/R^2
T1-T2=50A

SOLVE THE EQ AND GET THE ANS
PLZ RATE ME IF U FIND ME USEFUL

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