How Many Revolutions Does a Football Make in a Perfect Spiral Pass?

[leezak]

A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through each end of the ball. Suppose the ball spins at 7.04 rev/s. In addition, the ball is thrown with a linear speed of 21.6 m/s at an angle of 47.0° with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

I translated 47.0 degrees into radians and 7.04 into radians per sec and then used the rotational kinematics equation angle theta = 1/2 (omega initial - omega) * t, and then i multiplied t by 7.04 rev/sec... what am i doing wrong?? help! thanks

 

[Doc Al]

Start by figuring out how long the football is in the air.

 

[quicknote]

It seems like you are on the right track with using the rotational kinematics equation. However, it is important to note that in this scenario, the initial and final angular velocities are the same (since the ball is caught at the same height it was thrown), so the equation becomes theta = omega * t.

To find the time, t, you can use the kinematic equation for projectile motion: y = y0 + v0t + 1/2at^2. In this case, y0 and y are the same (height of the throw and catch), v0 is the initial velocity (21.6 m/s), and a is the acceleration due to gravity (9.8 m/s^2). Solving for t, we get t = 2v0sin(theta)/a (since y0 = y and a = -g).

Plugging in the values, we get t = 2(21.6 m/s)sin(47.0 deg)/9.8 m/s^2 = 3.45 seconds.

Now, to find the number of revolutions, we can use the equation omega = 2piN/t, where N is the number of revolutions. Rearranging for N, we get N = omega*t/2pi.

Plugging in the values, we get N = (7.04 rev/s)(3.45 s)/2pi = 12.3 revolutions.

Therefore, the ball has made approximately 12.3 revolutions while in the air.