A bowling ball of mass 7.20 kgkg and radius 10.0 cmcm rolls without slipping down a lane at 3.10 m/sm/s .Part ACalculate its total kinetic energy.Express your answer using three significant figures and include the appropriate units.
Question
A bowling ball of mass 7.20 kgkg and radius 10.0 cmcm rolls without slipping down a lane at 3.10 m/sm/s .Part ACalculate its total kinetic energy.Express your answer using three significant figures and include the appropriate units.
Solution
The total kinetic energy of the bowling ball is the sum of its translational kinetic energy and its rotational kinetic energy.
The formula for translational kinetic energy is (1/2)mv^2, where m is the mass and v is the velocity.
The formula for rotational kinetic energy for a sphere is (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
For a sphere, the moment of inertia, I, is (2/5)mr^2, where m is the mass and r is the radius.
The angular velocity, ω, can be found from the linear velocity, v, using the relationship v = rω. So, ω = v/r.
Let's calculate:
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Translational kinetic energy: K.E_trans = (1/2)mv^2 = 0.5 * 7.20 kg * (3.10 m/s)^2 = 34.344 J
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Rotational kinetic energy: First, calculate I: I = (2/5)mr^2 = 2/5 * 7.20 kg * (0.10 m)^2 = 0.00576 kg*m^2
Then, calculate ω: ω = v/r = 3.10 m/s / 0.10 m = 31.0 s^-1
Finally, calculate K.E_rot: K.E_rot = (1/2)Iω^2 = 0.5 * 0.00576 kg*m^2 * (31.0 s^-1)^2 = 27.716 J
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Total kinetic energy: K.E_total = K.E_trans + K.E_rot = 34.344 J + 27.716 J = 62.060 J
So, the total kinetic energy of the bowling ball is approximately 62.1 J.
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