Asked by Michael
A girl on a merry-go-round platform holds a pendulum in her hand. The pendulum is 3.7 m from the rotation axis of the platform. The rotational speed of the platform is 0.025 rev/s. It is found that the pendulum hangs at an angle è to the vertical. Find è.
Answers
Answered by
drwls
The angular speed of the platform is
w = (0.025 rev/s)*(2 pi) rad/rev
= 0.157 rad/s
The centripetal acceleration of the pendulum is
a_c = R*w^2 = 0.0913 m/s^2
The angle theta is
tan^-1(a_c/g) = 0.918 degrees
w = (0.025 rev/s)*(2 pi) rad/rev
= 0.157 rad/s
The centripetal acceleration of the pendulum is
a_c = R*w^2 = 0.0913 m/s^2
The angle theta is
tan^-1(a_c/g) = 0.918 degrees
Answered by
Anonymous
a girl on a rotating platform holds a pendulum in her hand. The pendulum is
at a radius of 6.0 m from the center of the platform. The rotational speed of the platform is 0.020 rev/s.
It is found that the pendulum hangs at an angle θ to the vertical, as shown. Find θ.
at a radius of 6.0 m from the center of the platform. The rotational speed of the platform is 0.020 rev/s.
It is found that the pendulum hangs at an angle θ to the vertical, as shown. Find θ.
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