a square loop of wire side 3.0 cm carries 3.0 A of current. A uniform magnetic field of magnitude 0.67 T makes an angle of 37 de
grees with the plane of the loop. What is the magnitude of the torque on the loop
1 answer:
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Answer:
The new angular speed of the merry-go-round is 8.31 rev/min.
Explanation:
Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):
(1)
The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:
(2)
with I the moment of inertia and ωi the initial angular speed of the merry-go-round
The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):
(3)
The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:
(4)
with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:
(5)
(note that the angular speed is the same as the merry-go-round)
using (5) on (4), and (4) on (3):
(6)
By (5) and (2) on (1):
Solving for ωf (12.0 rev/min = 1.26 rad/s):