Two astronaut, as shown in the figure, each having a mass of 62.0 kg, are connected by a 12.00 m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.00 m/s. Treating the astronauts as particles, calculate (b) the rotational energy of the system.
To calculate the rotational energy of the system, we need to find the moment of inertia and the angular velocity.
1. Moment of Inertia:
The astronauts can be approximated as point masses since their size is negligible compared to the length of the rope. The moment of inertia for two point masses is given by the equation:
I = m1 * r1^2 + m2 * r2^2
where m1 and m2 are the masses of the astronauts and r1 and r2 are the distances of the point masses from the center of rotation.
In this case, the astronaut masses are equal, so m1 = m2 = 62.0 kg. The distance from the center of rotation to each astronaut is half the length of the rope, which is 12.00 m / 2 = 6.00 m. Therefore, the moment of inertia for the system is:
I = 2 * (62.0 kg) * (6.00 m)^2
2. Angular velocity:
The angular velocity is the rate at which the system rotates in circles. It can be calculated using the formula:
ω = v / r
where v is the linear velocity and r is the distance from the center of rotation.
In this case, the linear velocity is 5.00 m/s, and the distance from the center of rotation to each astronaut is 6.00 m.
Now, we have all the necessary values to calculate the rotational energy.
3. Rotational Energy:
The rotational energy is given by the equation:
E = (1/2) * I * ω^2
where E is the rotational energy, I is the moment of inertia, and ω is the angular velocity.
Substituting the known values, we have:
E = (1/2) * [2 * (62.0 kg) * (6.00 m)^2] * [5.00 m/s / 6.00 m]^2
Solving this expression will give us the rotational energy of the system.