Two turtles (each 2 kg) sit on an approximately massless branch in the water. They sit on opposite sides of the branch's center, each 3 m away.
The branch (10 m long) is spinning at 2 rad/s about its center (nearly without friction).
Now, if both turtles walk to the ends of the log, then the new angular speed of the log+turtles system would be:
rad/s
do you use conservation of torque? what can be set equal?
<<do you use conservation of torque?>>
There is no torque on the branch. Angular momentum about the center of the branch is conserved.
They each move from 3 to 5 m from the center of the branch.
Angular velocity w decreases by a ratio 3/5 to keep the angular momentum constant
To solve this problem, we can use the principle of conservation of angular momentum. The angular momentum of a system remains constant unless an external torque is applied to it.
In this case, we have two turtles sitting on an approximately massless branch in the water, which is spinning at 2 rad/s about its center. The turtles are initially 3 m away from the center on opposite sides.
When both turtles walk to the ends of the log, their total moment of inertia will change. However, the conservation of angular momentum states that the initial angular momentum of the system must be equal to the final angular momentum of the system.
We can set the initial angular momentum equal to the final angular momentum and solve for the new angular speed of the log+turtles system.
Let's denote the mass of each turtle as m, the initial angular speed of the log as ω1, the final angular speed of the log+turtles system as ω2, and the length of the log as L.
The initial angular momentum (L_initial) is given by:
L_initial = (m × (3 m)^2) × ω1
The final angular momentum (L_final) is given by:
L_final = (2m × (L/2)^2) × ω2
Setting the initial angular momentum equal to the final angular momentum, we have:
(m × (3 m)^2) × ω1 = (2m × (L/2)^2) × ω2
Simplifying this equation, we get:
9 m^2 × ω1 = (1/2) m × (L^2) × ω2
Canceling out the mass (m) on both sides, we get:
9 m × ω1 = (1/2) m × (L^2) × ω2
Simplifying further, we obtain:
9 ω1 = (1/2) (L^2) × ω2
To find the new angular speed (ω2), we can rearrange the equation as follows:
ω2 = (9/(1/2) (L^2)) × ω1
Substituting the given values, we have:
ω2 = (9/(1/2) (10 m^2)) × 2 rad/s
Simplifying, we find:
ω2 = 18 rad/s
Therefore, the new angular speed of the log+turtles system, when both turtles walk to the ends of the log, would be 18 rad/s.
In conclusion, in order to solve this problem, we used the conservation of angular momentum principle and set the initial angular momentum equal to the final angular momentum to find the new angular speed of the log+turtles system.