The angular position of a point on the rim of a rotating wheel is given by = 6.0

Question

The angular position of a point on the rim of a rotating wheel is given by = 6.0t 1.0t2 + t3, where is in radians and t is in seconds.

(a) What is the angular velocity at t = 4.0 s? ______rad/s

(b) What is the angular velocity at t = 6.0 s? _______rad/s

(c) What is the average angular acceleration for the time interval that begins at t = 4.0 s and ends at t = 6.0 s?_ rad/s2

(d) What is the instantaneous angular acceleration at the beginning of this time interval? rad/s2 (e) What is the instantaneous angular acceleration at the end of this time interval? _____rad/s2

Explanation / Answer

Angular velocity is given by d/dt

d/dt = 6-2t+3t^2

a)

At t =2 , d/dt = 6-4+12 =14 rad/sec

b)

at t= 6 sec, d/dt = 6-12+108 =102 rad/sec

c)d2/dt = -2+6t

d2/dt = -2+6(6-4) = 10 rad/sec

d)

22 rad/sec

34 rad/sec

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