Question

A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18* to the horizontal, The cylinder has rolled 2.2m down the plane Draw sketch and calculate the change in height h of the cylinder? Using conservation of energy, derive an expression for the Iinear velocity of the cylinder in terms of the change in height h of the cylinder and calculate the linear velocity. What fraction of potential energy lost was converted to rotational kinetic energy?

Answer 1

a ) The change in height h of the cylinder = 0.68 m

b ) Expression for the linear velocity of the cylinder = 2.99 m / s

c ) Fraction of potential energy converted to rotational KE = 0.33

a ) The change in height h of the cylinder,

θ = 18°

Since it forms a right angled triangle, using trigonometric ratios

sin 18° = h / 2.2

h = 2.2 * 0.31

h = 0.68 m

b ) Expression for the linear velocity of the cylinder,

According to law of conservation of energy,

m g h + 0 = 0 + [ 1 / 2 m v² + 1 / 2 I ω² ]

m g h = 1 / 2 m v² + 1 / 2 ( 1 / 2 m r² ) ( v / r )²

m g h = ( 1 / 2 + 1 / 4 ) m v²

v² = 4 g h / 3

v² = 4 * 9.8 * 0.68 / 3

v² = 8.91

v = 2.99 m / s

c ) Potential energy converted to rotational kinetic energy,

Rotational kinetic energy, E1 = 1 / 2 m v² + 1 / 2 I ω²

E1 = 1 / 2 ( 1 / 2 m r² ) ( v / r )²

E1 = 1 / 4 m v²

Potential energy, E2 = m g h

E1 / E2 = 1 / 4 m v² / m g h

E1 / E2 = v² / 4 g h

E1 / E2 = 2.99² / ( 4 * 9.8 * 0.68 )

E1 / E2 = 8.91 / 26.66

E1 / E2 = 0.33

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