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September 20, 2014

September 20, 2014

Posted by **James** on Monday, June 18, 2012 at 6:39pm.

What is the speed of the solid ball at the bottom of the incline?

The answer is 3.96 m/s. How do I get this answer? At first I thought I could use E=1/2mv^2; however, I do not have the total amount of energy, so I cannot use this equation. Any help is greatly appreciated.

- Physics -
**Elena**, Tuesday, June 19, 2012 at 10:04amFor each object

PE → KE

h =s•sinα = 2•sin34º =1.12 m

Solid ball I = 2•m•R²/5, ω=v/R

PE =KE = KE (translational) + KE (rotational)

m•g•h = m•v²/2 + I•ω²/2 = m•v²/2 + 2•m•R²•v²/2•5R² =

= m•v²/2 + m• v²/ 5 =0.7•m• v².

v = sqrt(g•h/0.7) =sqrt(9.8•1.12/0.7) =3.96 m/s.

Hollow ball I = 2•m•R²/3 , ω=v/R

m•g•h = m•v²/2 + I•ω²/2 = m•v²/2 + 2•m•R²•v²/2•3R² =

= m•v²/2 + m• v²/ 3 =5•m• v²/6.

Ice cube

m•g•h = m•v²/2

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