The angular position of a point on the rim of a rotating wheel is given by = 4.5
ID: 1359469 • Letter: T
Question
The angular position of a point on the rim of a rotating wheel is given by = 4.5 t + (-8.0)t2 + (1.0)t3, where is in radians and t is given in seconds.
What is the angular velocity at t = 2.0 s?
What is the angular velociy at t = 4.0 s?
What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s?
What is the instantaneous angular acceleration at the beginning of this time interval?What is the instantaneous angular accelerations at the end of this time interval?
Explanation / Answer
Here ,
theta = 4.5 t - 8 t^2 + t^3
w = dtheta/dt
w = d/dt(4.5 t - 8 t^2 + t^3)
w = 4.5 - 16 * t + 3 t^2
at t = 2 s
w = 4.5 - 16 * 2 + 3 * 2^2
w = -15.5 rad/s
angular speed at t = 2 s is -15.5 rad/s
at t = 4 s
w = 4.5 - 16 * 4 + 3 * 4^2
w = -11.5 rad/s
the angular speed at t = 4 s is -11.5 rad/s
average angular acceleration = -11.5 -(-15.5) /(2)
average angular acceleration = 2 rad/s^2
the average angular acceleration is 2 rad/s^2
as a = dw/dt
a = -16 + 6 t
at t = 2 s
a = -16 + 6 * 2 = -4 rad/s^2
angular acceleration at t = 2 s sis - 4 rad/s^2
at t = 4 s
a = -16 + 6 * 4
a = 8 rad/s^2
the angular acceleration at t = 4 s is 8 rad/s^2
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