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A hollow cylinder is given a velocity of 5.3 m/s and rolls up an incline to a height of 2.87 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high (in m) does it roll up the incline

Respuesta :

Answer:

E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2

E = 1/2 M V^2 (1 + I / (M R^2))

For a cylinder I = M R^2

For a sphere I = 2/3 M R^2

E(cylinder) = 1 + 1 = 2       omitting the 1/2 M V^2

E(sphere) = 1 + 2/3 = 1.67

E(cylinder) / E(sphere) = 2 / 1.67 = 1.2

The cylinder initially has 1.20 the energy of the sphere

The PE attained is proportional to the initial KE

H(sphere) = 2.87 / 1/2 = 2.40 m    since it has less initial KE

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