1)The angular position of a point on the rim of a rotating wheel is given by = 5
ID: 1453116 • Letter: 1
Question
1)The angular position of a point on the rim of a rotating wheel is given by = 5.5 t + (-5.0)t2 + +(1.7)t3, where is in radians if t is given in seconds. What is the angular velocity at t = 2.0 s? Submit Answer Tries 0/8
What is the angular velociy at t = 4.0 s? Submit Answer Tries 0/8
What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? Submit Answer Tries 0/8
What is the instantaneous angular acceleration at the beginning of this time interval? Submit Answer Tries 0/8
What are the instantaneous angular accelerations at the end of this time interval?
Explanation / Answer
given
= 5.5 t + (-5.0)t2 + +(1.7)t3
the angular velocity at t = 2.0 s
w 1= d /dt = d/dt (5.5 t + (-5.0)t2 + +(1.7)t3)
= 5.5-10t+5.14 t^2
=5.5-10(2) +5.14(2)^2
=6.06 rad/s
at t= 4.0 s
w2= 5.5-10t+5.14 t^2
=5.5-10(4) +5.14(4)^2
=47.74 rad/s
(c)
the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s
alpha = w2- w1/ t2- t1 = 47.74 -6.06/4-2 =20.84 rad/s^2
(d)
the instantaneous angular acceleration at the beginning of this time interval
alpha = d^2/dt^2 =-10+10.28 t
t = 2 s
alpha = -10+ 10.28 (2) = 10.56 rad/s^2
t = 4 s
alpha = -10+ 10.28 (4) = 31.12 rad/s^2
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