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A uniform, solid sphere of radius 5.50 cm and mass 5.00 kg starts with a purely translational speed of 2.00 m/s at the top of an
inclined plane. The
surface of the incline is 2.00 m long, and is tilted at an angle of 29.0° with respect to the horizontal. The sphere rolls without slipping
down the incline.
Calculate the sphere's final translational speed v2 at the
bottom of the ramp.

Answer :

MathPhys

Answer:

4.05 m/s

Explanation:

Energy is conserved, so the sum of the sphere's initial potential energy and initial translational kinetic energy must equal the sum of the sphere's final translational kinetic energy and final rotational kinetic energy.

PE₀ + KE₀ = KE + RE

mgh + ½ mu² = ½ mv² + ½ Iω²

Moment of inertia for a sphere is I = ⅖ mr². For rolling without slipping, v = ωr.

mgh + ½ mu² = ½ mv² + ½ (⅖ mr²) ω²

mgh + ½ mu² = ½ mv² + ⅕ mv²

mgh + ½ mu² = ⁷/₁₀ mv²

gh + ½ u² = ⁷/₁₀ v²

Solve for v:

v² = ¹⁰/₇ (gh  + ½ u²)

v² = ¹⁰/₇ [ (9.8 m/s) (2.00 m sin 29.0°) + ½ (2.00 m/s)² ]

v = 4.05 m/s

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