IIT JEE , AIEEE Forum - Physics , Mechanics

sahilgupta_iit

A ring of mass m and radius r is rolling on a frictionless surface with the velocity of centre of mass v_o. Particles of mass m, 2m and 3m are rigidly attached to it as shown in the figure.

(1) Find the Kinetic Energy of the system

(2) The velocity of centre of mass of the system

rohitkuruvila

is the answer to the second part (4/3)v₀

anu08

ke of da system is
translational ke of ring+
rotational ke of ring+rotational ke of all da three masses as dey do not hav translational motion

akku

mag of velocity of m=(2)^0.5

mag of velocity of 2m=2vo

mag of velocity of 3m=(2)^0.5

total KE=TRANSLATIONAL KE OF m+TRANSLATIONAL KE OF 2m+TRANSLATIONAL KE OF 3m+TRANSLATIONAL KE OF RING(1/2*mvo^2)+ROTATIONAL KE OF RING(1/2*mR^2*Vo^2/R^2)

=9mvo^2

akku

xcm =3mr-mr/7m=2/7*r
ycm=2mr/7m=2/7*r
rcm=2/7*r(i+j)
velocity of com=voi+2/7voi-2/7voj=9/7voi-2/7voj ans

varun.tinkle

ASSUMING THE PARTICLES R ATTACHED AFTER MOCEMENT BEGINS
THEN
Iw=(I+2MR^2)W
MR^2W=3MR^2W1
W1=MR^2W/3MR^2
W1=W/3
SO FROM THIS WE CAN FIND OIT THE K.E

varun.tinkle

THE TOTA VELOCITY OF THE CENTRE OF THE RING
IS V (=WR)
V=WR/3
THE VELOCITY OF THE MASS 2M
IS V+WR
=2WR/3
VELOCITY OF THE MASS 3M IS INNDER ROOT (V)^2+(WR)^2
=(2WR)^2 =
AND THE VELCOITY OF THE CENTRE OF MASS OF THE COM O F THR SYSTEM
IS M1V+M2V+M3V/M1+M2+M3
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