contestada

If a 0.15-kg baseball with a radius of 3.7 cm is thrown with a linear speed of 41 m/s and an angular speed of 45 rad/s , how much of its kinetic energy is translational energy? assume the ball is a uniform, solid sphere.

Respuesta :

Total Kinetic energy of the ball is given by

[tex]KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2[/tex]

[tex]KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} mv^2[/tex]

[tex]KE = \frac{1}{2}(\frac{2}{5}*0.15*0.037^2)*45^2 +\frac{1}{2} *0.15*41^2[/tex]

[tex]KE = 0.083 + 126[/tex]

[tex]KE = 126.083 J[/tex]

Total translational KE of the ball is given by

[tex]KE_T = \frac{1}{2}mv^2[/tex]

[tex]KE_T = 126 J[/tex]

Fractional part of translational KE is given by

[tex]fraction = \frac{KE_T}{KE}[/tex]

[tex]fraction = \frac{126}{126.083}[/tex]

[tex]fraction = 0.99[/tex]

So translational KE is 99% of total KE of the ball

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