IIT JEE , AIEEE Forum - Physics , Mechanics

me_coolguy_iit_aspirant

a good question from hcv.......

the question goes in this way!!!!!!

The descending pulley has a radius 20cm. and MOI=0.20kg-m^2.The fixed pulley is light and the horizontal plane frictionless.Find the acceleration of the block if its mass is = 1.0 kg.

FOR FIGURE PLEASE REFER THE BOOK H.C VERMA::VOLUME-1 ON PAGE NUMBER-197, QUESTION NUMBER 37!!!!

PLEASE REPLY!!!!!

me_coolguy_iit_aspirant

REPLY PLZ!!!!

me_coolguy_iit_aspirant

no replies??????????

wasp goiitians?????

me_coolguy_iit_aspirant

wasp???

no one is trying to solve this?????????

me_coolguy_iit_aspirant

any reply plz........

biki

Consider the length of string method (example illustrated in page 73, q no. 6 of HC Verma vol.1) to find out the accl^n of the bodies
and you will get accl^n of body = 2 x accl^n of pulley(a)
So accl^n of body = 2a
Thus taking T1 as tension in the longer part of the string (attached to the block) and T2 as tension in the shorter part (fixed to the wall), we find that as the pulley(a disc or a cylinder) descends rotating about its axis, the torque acting on it = T2-T1
Thus (T2-T1) x radius of pulley(r) = Moment of Inertia(I) x angular accl^n
=> (T2-T1) x r = Ia/r2 ...{angular accl^n= a/r}
=> a = (T2-T1)r2/I_____(1)
Now write the force equations of the two bodies (block and pulley)....
T1 = 1 x 2a....(2) {mass of body = 1kg}
and 10 x 9.8 - T1 - T2 = 10a....(3) {10kg = mass of pulley as calculated from the relation I = Mr2/2}
Thereby from the equations (2) & (3) obtain the value of T2-T1
you will get it as T2-T1 = 98 - 14a
Substitute the value of T2-T1 in equation (1) and you will get an equation with 'a' as variable...
(1) => a = (98 - 14a)r2/I
Put the values of r and I
and you will get a = 5 m/s2
So the accl^n of the pulley = 5 m/s2
and thus accl^n of the block = 2 x accl^n of the pulley
= 2 x 5
= 10 m/s2

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