## RaphaelFilgueiras Group Title A spherical uniform shell rotates in a vertical axis without friction.One string without mass passing through sphere equator and a sheave has a small object at its end.what is the velocity of object,when he down a distance H. 2 years ago 2 years ago

1. RaphaelFilgueiras Group Title

2. RaphaelFilgueiras Group Title

Use work energy theorem

3. celos Group Title

rotational kinetic energy of spherical shell + rotational K.E. of pulley + K.E. of object = change in P.E. of object

4. celos Group Title

since string is not stretchable: velocity of object = angular speed of shell* Radius of shell = angular speed of pulley * radius of pulley. now put the known values and get the answer.

5. CarlosGP Group Title

$1/2mv^2+1/2I_{sphere}\omega_{sphere}^2+1/2I_{pulley}\omega_{pulley}^2=mgH$

6. RaphaelFilgueiras Group Title

@CarlosGP ,but we dont have w,and v yet

7. celos Group Title

w= v/r

8. celos Group Title

you will get a equation in one variable and will be able to find solution also

9. RaphaelFilgueiras Group Title

@carlos i get it i will try

10. CarlosGP Group Title

$I_{sphere}=2/5MR^2$ then, for the sphere $1/2(2/5MR^2)w_{sphere}^2=1/5Mv^2$ and for the pulley: $1/2Iw_{pulley}^2=1/2I(v/r)^2$ with the data given in the drawing: $mgH=v^2[m/2+M/5+I/(2r^2)]$

11. RaphaelFilgueiras Group Title

Isphere=2/3MR²

12. RaphaelFilgueiras Group Title

@CarlosGP thanks

13. CarlosGP Group Title

@RaphaelFilgueiras you are welcome. Isphere is 2/5MR^2

14. RaphaelFilgueiras Group Title

@CarlosGP it's just a shell

15. CarlosGP Group Title

You are right: I=2/3MR^2 for the sphere