- #1

SiRiVeon

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- Homework Statement
- What is the minimum height of the triangle for which if the black ball falls through the inclined

plane, it will be able to make a full cycle around the circle at the bottom of the inclined plane of

radius R?

- Relevant Equations
- mgh = (1/2)mv² + (1/2)Iω² , = (2/5)mr² , v = ωr , Centripetal force = mv²/r

Hello, this question may seem weird but I really need help on this.

To bring the formula for the height

For a sphere falling through an inclined plane, the equation to satisfy it's conservation of energy is

But in the given question, it says that the ball makes a full cycle around the circle of radius

To bring the formula for the height

**h**of the triangle above, I have to create a relation between potential and kinetic energies of the black ball with mass**m**(I can't find any other methods than this).For a sphere falling through an inclined plane, the equation to satisfy it's conservation of energy is

**mgh = (1/2)mv² + (1/2)**where*I*ω²**v**is the linear velocity of the ball,**g**is the acceleration due to gravity,**is the rotational velocity of the ball and****ω****is the moment of inertia of the ball which can be expressed by***I***=***I***kmr**where**²****k****= 2/5**for sphere of radius**r**. By putting**and expression of****ω = v/r****into the first equation, I get***I***mgh = (1/2)mv² + (1/5)mv²**.But in the given question, it says that the ball makes a full cycle around the circle of radius

**R**at the bottom of the inclined plane. So what should be the conservation of energy in this case? If the centripetal force to move around that circle is**mv**then will it be**²/R****mgh = (1/2)mv² + (1/5)mv² +**for the whole scenario or will it be**mv****²/R****mv**or none of this?**²/R =****(1/2)mv² + (1/5)mv²**