A softball player swings a bat, accelerating it from rest to 2.6 rev/s in a time of 0.20 s. Approximate the bat as a 0.90-kg uniform rod of length 0.95 m, and compute the torque the player applies to one end of it.
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This is Giancoli Answers with Mr. Dychko. The torque needed to accelerate the bat is equal to its moment of inertia times its angular acceleration. We assume the bat is a uniform rod with an axis of rotation at one end where the batter's hands are and so the formula to calculate its moment of inertia is one-third times its mass times its length squared. And then its angular acceleration is the final angular velocity minus the initial divided by time and since the initial angular velocity is zero—it starts from rest— we can rewrite this as ml squared ω f over 3t. And so we plug in numbers and paying really careful attention to the units here; they are being tricky here by giving us revolutions per second, usually they say, you know, revolutions per minute but in this case, it's 2.6 revolutions per second and so we convert that into radians per second by multiplying by 2π radians per revolution and times by the length 0.95 meters squared times the mass 0.90 kilograms divided by 3 times 0.20 seconds and we get 22 newton meters must be the net torque and there we go!
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