The angular position of a point on the rim of a rotating wheel isgiven by = 8.0

Question

The angular position of a point on the rim of a rotating wheel isgiven by = 8.0t- 2.0t2 +t3, where is in radians andt is in seconds. (a) What is the angular velocity at t= 4 s?
1 rad/s

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

(c) What is the average angular acceleration for the time intervalthat begins at t = 4 s and endsat t = 6.0 s?
3 rad/s2

(d) What is the instantaneous angular acceleration at the beginningof this time interval?
4 rad/s2

(e)What is the instantaneous angular acceleration at the end ofthis time interval?
5 rad/s2
(a) What is the angular velocity at t= 4 s?
1 rad/s

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

(c) What is the average angular acceleration for the time intervalthat begins at t = 4 s and endsat t = 6.0 s?
3 rad/s2

(d) What is the instantaneous angular acceleration at the beginningof this time interval?
4 rad/s2

(e)What is the instantaneous angular acceleration at the end ofthis time interval?
5 rad/s2

Explanation / Answer

= 8.0t -2.0t2 +t3, a) (t) = d/dt = 8.0 - 4.0t + 3t2 at t = 4.0 s, (4) = 40 rad/s b) at t = 6.0 s (6) = 92 rad/s c) = [(6) - (4)]/(6 - 4) = 26rad/s2 d) (t) = d(t)/dt = -4.0 + 6t (4) = -4 + 24 = 20 rad/s2 e) (6) -4.0 + 36 = 32 rad/s2

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