Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.30 mm high. Assume it starts from rest and rolls without slipping.

9.77 m/s

Explanation:

Since the cylinder is rolling, it will have both translational and rotational kinetic energy.

From conservation of energy, its initial mechanical energy equals its final mechanical energy.

So, K' + U' = K + U where K and U are the initial kinetic and potential energies respectively and K' and U' are the final kinetic and potential energies respectively.

So, Since the cylinder starts from rest, its initial kinetic energy K = 0, its initial potential energy = mgh where m = mass of cylinder, g = acceleration due to gravity = 9.8 m/s² and h = height of incline = 7.30 mm = 7.3 × 10⁻³ m . Its final kinetic energy K' = 1/2Iω² + 1/2mv² where I = moment of inertia of cylinder = 1/2mr² where r = radius of cylinder and v = translational velocity of cylinder. Its final potential energy U' = 0. Substituting these into the equation, we have  

K' + U' = K + U

1/2Iω² + 1/2mv² + 0 = 0 + mgh

1/2(1/2mr²)ω² + 1/2mv² = mgh

1/4mr²ω² + 1/2mv² = mgh          

1/4m(rω)² + 1/2mv² = mgh    v = rω

1/4mv² + 1/2mv² = mgh

3/4mv² = mgh

v² = 4gh/3

v = √(4gh/3)

v = √(4 × 9.8 m/s² × 7.3 × 10⁻³ m/3)

v = (√286.16/3) m/s

v = (√95.39) m/s

v = 9.77 m/s

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