Need help with a qn on rotational motion.

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SUMMARY

The discussion focuses on calculating the minimum velocity required for a hoop of radius r and mass m to roll up a step of height h without slipping. The user attempted to apply conservation of energy principles, leading to the equation v^2 = 3/2 gh, which was identified as incorrect. The correct formula for the minimum velocity is derived as v = 2r(gh)^(1/2)/(2r-h). The user notes that if the height h equals the radius r, the hoop cannot ascend the step regardless of its speed.

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A hoop pf radius r and mass m is rolling without slipping with velocity v towards a step of height h on a horizontal surface. Assume that it does not rebound and no slipping occur at the point of contact when the hoop roll up, what is the minimum velocity needed for the hoop to roll up?



I tried to use conservation of energy to solve:
Ek(transl.)i +EK(rotat.)i=mgh+(Torque.change in angle of rotation)
1/2 mv^2 + 1/2 I w^2 = mgh + T.ditre
...
1/2 mv^2 + 1/2 mr^2(v^2/r^2)= mgh + m(r^2).(angu. acele)(ditre)
v^2=3/2 gh

this is far from the correct answ of 2r(gh)^0.5/(2r-h)

Please help me
 
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The correct answer doesn't look quite right either. If h = r then the hoop won't make it up the step at any speed.

In any case, only the component of velocity perpendicular to a line joining the corner of the step and the center of the hoop can contribute to lifting the hoop.
 

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