Asked by JRD
The speed of a point on a rotating turntable,
which is 0.172 m from the center, changes at a
constant rate from rest to 0.531 m/s in 2.19 s.
At t1 = 1.9 s, find the magnitude of the
tangential acceleration.
Answer in units of m/s2.
which is 0.172 m from the center, changes at a
constant rate from rest to 0.531 m/s in 2.19 s.
At t1 = 1.9 s, find the magnitude of the
tangential acceleration.
Answer in units of m/s2.
Answers
Answered by
drwls
The angular acceleration rate is
alpha = (Angular velocity at 2.19 s)/2.19s
= (0.531/0.172)/2.19 = 1.41 rad/s^2
The tangential acceleration rate remains constant at R*alpha = 0.242 m/s^2 during the turntable acceleration. The 1.9 s time is not needed for the calculation, since alpha remains constant. Centripetal acceleration increases with the square of time.
alpha = (Angular velocity at 2.19 s)/2.19s
= (0.531/0.172)/2.19 = 1.41 rad/s^2
The tangential acceleration rate remains constant at R*alpha = 0.242 m/s^2 during the turntable acceleration. The 1.9 s time is not needed for the calculation, since alpha remains constant. Centripetal acceleration increases with the square of time.
Answered by
JRD
At t1 = 1.9 s, find the magnitude of the total
acceleration of the point.
Answer in units of m/s2.
acceleration of the point.
Answer in units of m/s2.
Answered by
drwls
I just did
Answered by
JRD
oh okay sorry I'm not the best at this and iv been super stuck >< thank you for the help
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