The angular position of a point on the rim of a rotating wheel is given by = 5.5

Question

The angular position of a point on the rim of a rotating wheel is given by = 5.5 t + (-3.5)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

given that

theta = 5.5t + (-3.5)*t^2 + (1.0)*t^3

we know that angular velocity

w = d(theta) /dt

w = d(5.5*t -3.5*t^2 + t ^3)/dt

w = 5.5 - 7*t + 3*t^2

part (a) w at    t= 2 sec

w( 2 ) = 5.5 - 7*2 + 3*4

w( 2) = 3.5 rad/s

part (b)

w at     t = 4s

w(4 ) = 5.5 - 7*4 + 3*16

w(4) = 25.5 rad/s

part(c)

average angular acceleration alpha is

ave alpha = delta w/ delta t

ave alpha for time interval t = 2 s to t = 4 s

ave alpha = (25.5 - 3.5 )/ 2

ave alpha = 22 / 2

ave alpha =11 rad / s^2

part(d)

alpha = dw / dt

alpha = - 7 +6*t

alpha at   t = 2 sec

alpha (2) = -7 +6*2

alpha (2) = 5 rad /sec^2

part(e)

alpha at     t = 4 sec

alpha (4 ) = -7 + 6*4

alpha (4) = 17 rad /sec^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.