A baton twirler throws a spinning baton directly upward. As it goes up and returns to the twirler's hand, the baton turns through five revolutions. Ignoring air resistance and assuming that the average angular speed of the baton is 1.70 rev/s, determine the height to which the center of the baton travels above the point of release.
in meters.
Okay. How do I figure out the distance from angular kinematic equations? What equations would I use? (Yes, I feel kind of stupid because this seems like a simple question, but I can't figure it out).
2006-11-03 13:29:45 · 2 answers · asked by Confused 1 in Science & Mathematics ➔ Physics
The baton made the "round trip" in 5 revolutions. Since it takes 1.70 rev for every second, total time=5/1.7=2.94s.
This means that it took half this time from the top of the rise till it reaches the twirler's hand or 2.94/2=1.47s.
We know that at the top of the rise the speed was 0, and as a freely falling body, it was accelerating down at 9.8m/s^2.
Use the formula: s=ut +1/2at^2 where s is the height, u the initial velocity at the top equal to 0, a the acceleration of gravity equal to
9.8m/s^2, and t the time of 1.47s.
Substitute known values and solve for s:
s=0*t+1/2*9.8*1.47^2
=10.6m
2006-11-03 16:45:12 · answer #1 · answered by tul b 3 · 0⤊ 0⤋
Fisrt use the angular part to determine total time spent in the air. Then use this time with the regular (linear) kinematic equations to solve for the height.
2006-11-03 14:45:46 · answer #2 · answered by Anonymous · 0⤊ 0⤋