Physics Question?

Question

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to 5.95 rev/s while rotating through one-half of a revolution. How much time does this maneuver take?

Answer 1

The data seems incomplete. I'll make some assumptions:

Her angular velocity increases in a linear manner from 3.00 to 5.95 rev/s while rotating through one-half of a revolution. So her average angular velocity during the maneuver is 4.97 rev/s. Then
1/2 rev * (1 sec/4.97 rev) = 0.1 s.

Answer 2

Let us suppose that the gymnast increases her angular velocity uniformly in time, that is her angular velocity is
W = W(0) + AT,
where T is time, W(0) is angular velocity (measured in rev/s) at the beginning of her maneuver and equals 3rev/s, and A means the angular acceleration, which we still did not calculate.

The angle of rotation dF for a small period of time dT will be:
dF = WdT = W(0)dT + ATdT
So, during the maneuver, the gymnast will spin according to the formula:
F = W(0)T + 0.5AT^2,
where T^2 means T multiplied by T, or second degree of T, and F means angle of rotation starting from the beginning of the maneuver at zero time T.
The full revolution will correspond to the angle F = 1(rev) (in numbers or reverses), so one half of revolution corresponds to angle F = 0.5(rev).
Let us now write the equation for the moment of time, when she ends her maneuver:
0.5(rev) = 3(rev/s)T(e) + 0.5AT(e)^2,
here T(e) is the time (in s) which we have to answer according to the question above.
But we still do not know the value of A. To find A, we shall write the second equation for
the angular velocity of the gymnast at the time when she ends her maneuver:
W(e) = W(0) + AT(e), and from this we shall derive the expression for A:
A = (W(e) - W(0))/T(e) = (5.95-3.00)(rev/s)/T(e) = 2.95(rev/s)/T(e)
Now we can insert this expression to the formula
0.5(rev) = 3(rev/s)T(e) + 0.5AT(e)^2, and thus receive
0.5(rev) = 3(rev/s)T(e) + (0.5x2.95(rev/s))T(e) = (3 + 1.475)(rev/s)T(e) =
= 4.475(rev/s)T(e).
It is simple to find T(e) from here:
T(e) = 0.5/4.475 = 0.11173 s

This is the answer.
It is also easy to obtain just finding the average angular velocity during the time of the maneuver as
(3 + 5.95)/2 = 4.475 rev/s,
and then just deviding one-half of the revolution to the average angular velocity:
0.5(rev)/4.475(rev/s) = 0.11173(s)
Physics is simple.

Answer 3

AP Physics is tough. If your math grades are at least a B then I would consider taking AP Physics. It will be tough for you unless you are really strong in math.

Answer 4

dU=w1*t+w’*t^2/2, w2=w1+w’*t, dU=pi=(1/2) rev, t is time for maneuver, w1=3*2pi rad/s, w2=5.95*2pi rad/s, w’ is her angular acceleration rad/s^2. thus w’=(w2-w1)/t & dU=w1*t+(w2-w1)*t/2, hence 2*dU=(w1+w2)*t, t=2*dU/(w1+w2)=1/(3+5.95)=0.11 s