Two astronaut, as shown in the figure, each having a mass of 88.0 kg, are connected by a 10.00 m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 6.00 m/s. By pulling on the rope, the astronauts shorten the distance between them to 5.00 m. (d) What are their new speeds?
To find the new speeds of the astronauts, we can use the principle of conservation of angular momentum. Angular momentum is defined as the product of an object’s moment of inertia (I) and its angular velocity (ω).
The moment of inertia of a point mass rotating in a circle about an axis through its center can be calculated using the equation I = m*r^2, where m is the mass of the object and r is the distance from the axis.
The initial angular momentum of the system is given by L_initial = I_initial * ω_initial, and the final angular momentum is given by L_final = I_final * ω_final.
Since no external torque acts on the system, the initial angular momentum will be equal to the final angular momentum. Thus, L_initial = L_final.
Now, let's calculate their initial and final angular momenta:
Initial angular momentum (L_initial):
Since the astronauts are moving in circles with a speed of 6.00 m/s and a distance of 10.00 m, the initial angular velocity (ω_initial) can be calculated using the equation v = ω * r, where v is the linear velocity and r is the distance.
ω_initial = v_initial / r_initial = 6.00 m/s / 10.00 m = 0.600 rad/s
The moment of inertia (I_initial) is given by the sum of the moments of inertia of the two astronauts:
I_initial = (m1 * r1^2) + (m2 * r2^2), where m1 and m2 are the masses of the astronauts, and r1 and r2 are their respective distances from the axis.
I_initial = (88.0 kg * (10.00 m / 2)^2) + (88.0 kg * (10.00 m / 2)^2) = 19280 kg·m²
L_initial = I_initial * ω_initial = 19280 kg·m² * 0.600 rad/s = 11568 kg·m²/s
Next, let's calculate their final angular momentum (L_final) after they shorten the distance between them to 5.00 m.
The new initial and final angular velocities (ω') can be calculated using the equation v' = ω' * r', where v' is the new linear velocity and r' is the new distance.
v' = 6.00 m/s (since their linear velocity remains the same)
r' = 5.00 m
ω' = v' / r' = 6.00 m/s / 5.00 m = 1.20 rad/s
The moment of inertia (I') is given by the sum of the moments of inertia of the two astronauts at the new distance:
I' = (m1 * r1'^2) + (m2 * r2'^2), where r1' = r2' = r' (since they are equidistant from the axis)
I' = (88.0 kg * (5.00 m / 2)^2) + (88.0 kg * (5.00 m / 2)^2) = 8800 kg·m²
Now, we can find the final angular momentum:
L_final = I' * ω' = 8800 kg·m² * 1.20 rad/s = 10560 kg·m²/s
Since angular momentum is conserved, we have L_initial = L_final:
11568 kg·m²/s = 10560 kg·m²/s
Therefore, the new speeds of the astronauts will be equal to their initial speeds, which is 6.00 m/s.