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A bowling ball has a mass of 3.7kg, a moment of inertia of 0.340992kg*m^2, and a radius of 0.48m. It rolls along the lane without slipping at a linear speed of 3.2m/s. What is the total kinetic energy of the rolling ball. Answer in units of J.

Respuesta :

egenriether
Kinetic energy due to linear motion:

[tex]E= \frac{1}{2}mv^2= \frac{1}{2}3.7(3.2)^2=18.944J[/tex]

The angular velocity is found by 

[tex]v=r \omega [/tex]

[tex]\omega = \frac{v}{r}= \frac{3.2}{0.48} =6.667rad/s[/tex]

Kinetic energy due to rotational motion:

[tex]E= \frac{1}{2}I \omega^2 = \frac{1}{2}0.340992(6.667)^2=7.5776J[/tex]
Then
[tex]E_{total}=18.944J+ 7.5776J=26.5216J[/tex]


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