A 50 kg skater moving at 5 m/s over frictionless ice grabs a rope stretched out 10m from a post and perpendicular to her path. She then moves in circles, spiraling inward as the rope wraps around the post. How fast is she moving when she is 2 meters from the post?
angular momentum must conserved.
OrigAngMomentum=finalangmomentum
I*wi=Iwf
M*ri^2*vi/ri=M*rf^2*Vf/rf
Vi*ri=Vf*rF
vf=5*10/2 m/s
To determine the skater's speed when she is 2 meters from the post, we can utilize the conservation of angular momentum.
First, let's consider the initial state of the skater. At this point, the skater is moving in a straight line with a constant velocity of 5 m/s. Since there is no external force acting on the skater in the direction perpendicular to her motion, her angular momentum is conserved.
The angular momentum (L) of the skater is given by the expression L = mvr, where m is the mass of the skater, v is her linear velocity, and r is the distance between the skater and the post. In this case, m = 50 kg, v = 5 m/s, and r = 10 m. Thus, the initial angular momentum of the skater is L_initial = 50 kg * 5 m/s * 10 m = 2500 kg·m²/s.
As the skater moves in circles and spirals inward, the distance between the skater and the post decreases. When the skater is 2 meters from the post, the new distance (r') is 2 m.
To find the final linear velocity (v') of the skater when she is 2 meters from the post, we can use the conservation of angular momentum by equating the initial angular momentum to the final angular momentum.
L_initial = L_final
Therefore, mvr = mv'r'
By substituting the given values, we obtain:
50 kg * 5 m/s * 10 m = 50 kg * v' * 2 m
Simplifying the equation, we find:
500 kg·m²/s = 100 kg·m/s * v'
Dividing both sides of the equation by 100 kg·m/s, we get:
v' = 5 m/s
Hence, the skater's speed when she is 2 meters from the post is 5 m/s.