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.

Question

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.

anonymous · Answer

attachment

anonymous · Answer

Use work energy theorem

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

w= v/r

anonymous · Answer

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

anonymous · Answer

@carlos i get it i will try

anonymous · Answer

$$I_{sphere}=2/5MR^2$$ 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:
$$mgH=v^2[m/2+M/5+I/(2r^2)]$$

anonymous · Answer

Isphere=2/3MR²

anonymous · Answer

@CarlosGP thanks

anonymous · Answer

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

anonymous · Answer

@CarlosGP it's just a shell

anonymous · Answer

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