A baseball pitcher loosens up his pitching arm. He tosses a 0.15-kg ball using only the rotation of his forearm, 0.26 m in length, to accelerate the ball. If the ball starts at rest and is released with a speed of 23 m/s in a time of 0.4 s, what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?Select one:a.3.3 N·mb.33.2 N·mc.22.4 N·md.221.2 N·me.2.2 N·m
Question
A baseball pitcher loosens up his pitching arm. He tosses a 0.15-kg ball using only the rotation of his forearm, 0.26 m in length, to accelerate the ball. If the ball starts at rest and is released with a speed of 23 m/s in a time of 0.4 s, what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?Select one:a.3.3 N·mb.33.2 N·mc.22.4 N·md.221.2 N·me.2.2 N·m
Solution
To solve this problem, we first need to find the angular acceleration. We know that angular acceleration (α) is equal to the change in angular velocity (Δω) divided by the change in time (Δt).
The angular velocity (ω) can be found by dividing the linear velocity (v) by the radius (r). In this case, the linear velocity is the speed of the ball when it is released (23 m/s) and the radius is the length of the pitcher's forearm (0.26 m).
So, ω = v / r = 23 m/s / 0.26 m = 88.46 rad/s.
Since the ball starts from rest, the initial angular velocity is 0. Therefore, the change in angular velocity is simply the final angular velocity, which is 88.46 rad/s.
Now we can find the angular acceleration: α = Δω / Δt = 88.46 rad/s / 0.4 s = 221.15 rad/s².
Finally, we can find the torque (τ) using the formula τ = Iα, where I is the moment of inertia. For a point mass, the moment of inertia is I = mr², where m is the mass of the ball (0.15 kg) and r is the radius (0.26 m).
So, I = 0.15 kg * (0.26 m)² = 0.01014 kg*m².
And τ = Iα = 0.01014 kgm² * 221.15 rad/s² = 2.24 Nm.
So, the correct answer is c. 2.24 N·m.
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