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A solid sphere rolls up a plane inclined at 45^(@)to the horizontal. If the speed of its centre of mass at the bottom of the plane is 5 m/s , find how far the sphere travels up the plane. |
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Answer» Solution :Data : V= 5 m/s ,` theta= 45^(@) , g= 9.8 ` m/`s^(2)` The total energy of the sphere at the bottom of the inclined plane is `E = 7/10 Mv^(2)` Where M is the mass of the sphere. In ROLLING up the plane through a vertical height h, it TRAVELS a distance L along the plane. Then. ` h - L SIN 45^(@) = L /sqrt2` By conservation of energy , Mgh = E ` Mg = L/sqrt2 = 7/10 Mv^(2)` ` L = (7sqrt2)/10 . v^(2)/g = ( 7 xx 1.414)/1 . ((5)^(2))/9.8 = ( 4.949 xx 5)/ 9.8` = `24.75/(0.8) = 2.526 m` The sphere travels 2.526 m up the inclined plane. |
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