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

AI Thread Summary
To determine how many revolutions a football makes during a perfect spiral pass, first calculate the time the ball is in the air using its launch speed and angle. The horizontal and vertical components of the throw must be analyzed to find the total flight time. Once the time of flight is established, multiply it by the spin rate of 7.04 revolutions per second to find the total revolutions. The discussion emphasizes the importance of correctly applying kinematic equations and understanding the relationship between linear and rotational motion. Accurate calculations will yield the total number of revolutions made by the football while airborne.
leezak
Messages
43
Reaction score
0
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
 
Physics news on Phys.org
Start by figuring out how long the football is in the air.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top