When throwing a ball, a pitcher exerts a torque of 150 N m for 0.15 seconds onto their arm before letting go of the ball. With an arm length of 0.9 m with inertia 0.6 kg m2, what is the speed of the ball just after it is released?
Paige, check your 4-6-11,4:23pm post.
i sill don't get it?
To calculate the speed of the ball just after it is released, we can use the principle of conservation of angular momentum. The equation for angular momentum is given by:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In this case, the pitcher exerts a torque of 150 Nm for 0.15 seconds. Torque is defined as the product of force and distance, so we can calculate the angular impulse:
τ = ΔL = IΔω
Given that the torque is 150 Nm, the time is 0.15 seconds, and the moment of inertia is 0.6 kg m^2, we can solve for Δω:
150 Nm = 0.6 kg m^2 * Δω
Δω = 150 Nm / 0.6 kg m^2 = 250 rad/s
The initial angular velocity can be calculated by dividing the angular impulse by the moment of inertia:
ω_initial = Δω / I = 250 rad/s / 0.6 kg m^2 = 416.67 rad/s
Since we know that the distance from the axis of rotation to the center of mass of the arm is 0.9 m, we can find the linear velocity v of the ball just after it is released using the formula:
v = r * ω
where r is the distance and ω is the angular velocity.
v = 0.9 m * 416.67 rad/s = 375 m/s
Therefore, the speed of the ball just after it is released is approximately 375 m/s.