- #1

Krushnaraj Pandya

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## Homework Statement

A metallic spherical shell of mass M and radius R is fully filled with water of mass M is set into pure pure rolling motion on a smooth horizonal surface (w clockwise, v to the right). Now water is freezed suddenly, and the ice sticks with the shell. What is-

a) moment of Inertia of shell about point of contact

b) angular momentum of water about point of contact

c) MI of shell+ice system about point of contact

d) MI of shell+water system about point of contact

e) new velocity of center of mass just after water freezes

## Homework Equations

All equations pertaining to rotational mechanics

## The Attempt at a Solution

a) since it is asked just for shell, I can forget about the water and write I(com) as 2/3mr^2 then apply parallel axes theorem to get answer as 5/3 Mr^2 (correct according to my textbook)

b) My intuition is that the water won't roll and will just have a straight translational velocity (since this question hasn't called fluid mechanics into question- we can forget about the drag of the liquid, internal friction between sphere and water etc.) and just write is as MvR (the correct answer as per the book) am I right?

c) I considered them as 2 separate spheres- since ice is solid, it'll behave as a normal solid sphere; so adding 2/3 + 2/5 + 2(parallel axis theorem) I got the correct answer 46/15 MR^2

d) here is where I'm getting confused, MI of a fluid is more since it tries to get away from the axis of rotation- I don't know how that'll come into play here and how I can calculate total MI

e) Angular momentum about point of contact is conserved therefore I(shell+water about COM)*w + 2Mvr=I(shell+ice) about com*w1 + 2Mv1r where w1 and v1 are new angular and linear velocitites, and v=rw is followed everywhere

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