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A solid, homogeneous sphere with of mass of M = 2.35 kg and a radius of R = 12.9 cm is resting at the top of an incline as shown in the figure.
The height of the incline is h = 1.29 m, and the angle of the incline is θ = 20.1°. The sphere is rolled over the edge very slowly. Then it rolls down to the bottom of the incline without slipping.
What is the total kinetic energy (translational plus rotational kinetic energy) of the sphere when it reaches the bottom of the incline?

A solid homogeneous sphere with of mass of M 235 kg and a radius of R 129 cm is resting at the top of an incline as shown in the figure The height of the inclin class=

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onyebuchinnaji

The total kinetic energy  of the sphere when it reaches the bottom of the incline is  44.57 J.

Total kinetic energy of the sphere

The total kinetic energy of the sphere = translational + rotational kinetic energy.

Translational kinetic energy

K.E = ¹/₂mv²

K.E(bottom) = P.E(top) = mgh

K.E = (2.35)(9.8)(1.29)

K.E = 29.71 J

Rotational kinetic energy

K.E(rot) = ¹/₂Iω²

K.E(rot) = ¹/₂I(v²/R²)

where;

  • I is rotational kinetic energy

I = ¹/₂MR²

K.E(rot) = ¹/₂( ¹/₂MR²)(v²/R²)

K.E(rot) = ¹/₄Mv²

K.E(rot) = ¹/₄M(2gh)

K.E(rot) = ¹/₂Mgh

K.E(rot) = ¹/₂(29.71) = 14.855 J

Total kinetic energy = 29.71 + 14.855 = 44.57 J

Learn more about kinetic energy here: https://brainly.com/question/25959744

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