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A uniform, solid sphere of radius 4.50 cm and mass 4.50 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 1.00 m long, and is tilted at an angle of 29.0 ∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed v 2 at the bottom of the ramp.

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Answer:

3.58 m/s

Explanation:

Data:

The plot can be drawn to present the information in the question.

Then the calculations:

mass = 4.50 kg

translational speed = 2 m/s

radius = 4.50 cm

angle  = 29.0 °

Energy at the top of the incline, [tex]E_{i}[/tex] is calculated as follows:

[tex]E_{i} = mgh\\ = mg(2.00) sin 32 + \frac{1}{2}mv_{i} ^{2} \\ = (4.50 kg) (9.81 m/s^{2}) (2.00m)sin32 + \frac{1}{2}(4.50 kg) (2.50 m/s)\\ = 52. 36 J[/tex]

The next item is the moment of inertia of the solid:

[tex]I = \frac{2}{5}r^{2}[/tex]

  = [tex]\frac{2}{5}(4.50)(0.045)^{2} \\= 0.003645[/tex]

Then the final velocity, [tex]v_{f}[/tex] is given by:

[tex]v_{f} = \frac{5}{7}*4+ (\frac{10}{7})* (9.81 * 0.714)^{2} \\ = 3.58 m/s[/tex]

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