1about which axis u have taken rotation
system shown in figure is in equilibrium. find the magnitude of net change in the string tension between two masses just after, one of the springs is cut. masses of both the blocks is same and equal to M. and spring constant of both the springs is 'k'
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14 Answers
106there are many things to be clarified:
(a) is the second block hanging in the middle?
(b) is the first block hanging with the end points attached to string?
Assuming, both this is how u shud solve:
(i) first it will make rotational motion.
So, taking torques,
2mg.L/2 = Iα (l = length of longer block)
Here I = ml2/3 + ml2/4 = 7ml2/12
So, α = mgl/7ml2/12 = 12g/7l
So, acom = α.l/2 = 6g/7......(i)
Now write force equation:
2mg-Fspring = 2ma
==> F=2mg/7
1chal agar end point k about rotation li hai to uske about inertia ml2/3 hona chaiye ?????
106ml2/3 to liya hoon. the only flaw which i can tink of is MI of the shorter block ..... whether it is M(l/2)2 ya nahin.....
106agar dono block lambe hote dono ka length same hota to answer mg/2 aata.....
1to ho sakta hai ye question galat ho. its from d.c pandey mechanics part 1............ nishant bro helpzzzzzzzz plzzzzzzzzzzzz
106which edition?, maybe i can see tabhi shaayad kuch dimaag mein a jaaye.
106according to the figure given, it has assumed that there is no rotational motion as the springs are very close almost at the COM..... so ur fig. is wrong...... as the spring is very close to COM, net torque will be zero. So, initially force exerted by each spring was 2mg/2 = mg
just after it is cut...................my mind stopped working...............nishant bhaiyya's help required or anyone who can solve this is welcome......