Asked by Malcom
A wheel spins on a horizontal axis, with an angular spped of 140rad/s and with its angular velocity pointing east. Determine the magnitude of the angular velocity after an angular acceleration of 35 rad/s^2, pointing 68 degress west of north, with a time of 4.6 seconds. I found that the magnitude of the angular velocity is 61 rad/s, but how do i determine the direction of its angular velocity west of north in degrees.
Answers
Answered by
drwls
Apply the vector equation
dw/dt = a
where w is the angular velocity and a is the angular acceleration.
a has components -21.55i (which points west) and 13.11 j (which points north).
w starts out at 140 i. After 4.6 seconds
w (t=4) = 40.87 i + 60.31 j
The magnitude does not agree with your answer. The direction tangent can be obtained from the ratio of the components.
Magnitude = 72.9
dw/dt = a
where w is the angular velocity and a is the angular acceleration.
a has components -21.55i (which points west) and 13.11 j (which points north).
w starts out at 140 i. After 4.6 seconds
w (t=4) = 40.87 i + 60.31 j
The magnitude does not agree with your answer. The direction tangent can be obtained from the ratio of the components.
Magnitude = 72.9
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