The other end of the string is fixed about a point. The ball rotates about this point at a frequency of 1 Hz/ While the ball is rotating, the string is reeled in until its length r = 3m. If the system is not otherwise disrupted in any way, what is the new frequency of rotation?
25/9 Hz
5/3 Hz
1 Hz
3/5 Hz
9/25 Hz
can you help out this hard and I've tried everything? Please
2007-03-28 03:48:13 · 3 answers · asked by Sesily E 1 in Science & Mathematics ➔ Physics
Angular momentum M conservation:
m v r = M0
v(r) = M0 / mr
Frequency:
f(r) = v(r)/2πr = M0 / 2πmr²
Therefore frequency is inversely proportional to r².
Answer: (5/3)² x 1Hz = 25/9 Hz
2007-03-28 04:58:17 · answer #1 · answered by Alexander 6 · 0⤊ 0⤋
Its to do with the conservation of angular momentum.
L = mvr
L is the angular momentum and must remain the same in both cases where r = 5 and r = 3. The mass of the ball is not going to change so the velocity must change.
The velocity after the string has been reeled in will be 5/3 of the original velocity.
The frequency is proportional to the velocity and so the answer is 5/3 Hz.
2007-03-28 04:03:02 · answer #2 · answered by Robin the Electrocuted 5 · 1⤊ 0⤋
Work out the angular velocity w. w = 2*PI*f
Use the fact that kinetic energy must be conserved in this system. (KE = 0.5*m*w^2/r)
Get a new angular velocity and hence convert back to frequency.
2007-03-28 03:57:35 · answer #3 · answered by dudara 4 · 1⤊ 0⤋