A baseball pitcher,loosening up his arm before a game,tosses a 0.140-kg ball using only the rotation of his forearm,0.290 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 24.0 m/s in a time of 0.450 s,what is the average angular acceleration of the arm and ball? A)3.13 rad/s² B)37.2 rad/s² C)15.5 rad/s² D)184 rad/s² E)25.7 rad/s²

Question

Quizplus · Accepted Answer

We can use the rotational kinematic equations to solve for the angular acceleration:
θ = θ0 + ω0t + 1/2αt^2
ω = ω0 + αt
where θ is the final angle (equal to 2π radians), θ0 is the initial angle (0 radians), ω is the final angular velocity (unknown), ω0 is the initial angular velocity (0 radians/s), t is the time (0.450 s), and α is the angular acceleration (unknown).

From the given information, we can calculate the final angular velocity:
v = rω
where v is the final tangential velocity (24 m/s), r is the length of the forearm (0.290 m), and ω is the final angular velocity (unknown):
ω = v/r = 24/0.290 = 82.8 rad/s

Using the second kinematic equation, we can solve for the angular acceleration:
ω = ω0 + αt
α = (ω - ω0)/t
α = (82.8 - 0)/(0.450)
α = 184 rad/s^2

Therefore, the average angular acceleration of the arm and ball is 184 rad/s^2. The correct answer is D.

A Baseball Pitcher,loosening Up His Arm Before a Game,tosses a 0.140-Kg