Sphere rotation expression help

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SUMMARY

The discussion focuses on deriving the total time of motion for a sphere of radius R projected up an inclined plane at an angle @ with an initial speed Vo and angular velocity Wo. The coefficient of friction is defined as tan@/7, and it is established that Vo > RWo. The solution involves calculating the time taken to achieve rolling conditions, resulting in an expression of (2Vo - 2WoR)gsin@ + 5, followed by a total stopping time of 14/3 seconds after rolling begins.

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Homework Statement


1) A sphere of radius R is projected up an inclined plane of angle @ with initial speed Vo and angular velocity Wo in a direction in which it can roll.The coefficient of friction is tan@/7 and it is given that Vo> RWo. Obtain an expression for the total time of motion of the sphere up the plane before it stops.

Comment on my working.

Homework Equations



I = 2/5mr^2
Impulse = delta mv = ft
Angular impulse = delta Iw = ftr ( r is perpendicular dist from axis of rotation)

The Attempt at a Solution



Concept:
Since Vo > RWo so the sphere will first try to perform rolling which is caused due to the presence of friction. Direction of friction will be backward along the incline so that the W increases and V decreases which eases the attainment of rolling. Once rolling has been achieved , friction changes direction(whoa!) because now to keep up with the decreasing V, it has to decrease W also.This will happen till the block is at an instantaneous rest.

Result:

First I found the time taken to reach rolling conditions which came out to be equal to (2Vo - 2WoR)gsin@ + 5 and the time after rolling till the time it comes to a stop came out to be 14/3 seconds (LOL?)
 
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