1mgh=1/2(Iw^2) +1/2mv^2
I=2/3mr^2
A uniform spherical shell of mass M rotates about vertical axis on frictionless bearings.A light cord passes around the equator of the shell over a masssless pulley and attached to block(mass m) which is otherwise free to fall under gravity.What is speed of object after it has fallen distance h from rest
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8 Answers
24I dont get answer form here......do u get √[2gh]/[1+(2M/3m)]???????
1yeah .........u get just take I as 2/3Mr^2 itz a hollw sphere
cheers~
12mgh=2/3Mr^2 w^2+mv^2
v=wr
so u get
2mgh=2/3Mv^2+mv^2
so nw u hav
2mgh/(2/3M+m)=v^2
dividin thruout by m u get
2gh/((2M/3m)+1) =v^2
then take d root
:)
24shit........silly mistake.......
i was writing 2mgh=2/3mv^2+mv^2 !!!!!!!!!!!!!!!!!!!!
anyways thanxxxxx