Chapter 10, Problem 006 The angular position of a point on the rim of a rotating

Question

Chapter 10, Problem 006 The angular position of a point on the rim of a rotating wheel is given by ?-1.73t-3.37t2 + 2.31t, where ? is in radians and t is in seconds. What are the angular velocities at (a) t = 3.19 s and (b) t = 9.91 s? (c) What is the average angular acceleration for the time interval that begins at t = 3.19 s and ends at t = 9.91 s? what are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval? (a) Number | (b) Number (c) Number (d) Number (e) Number Units rad/s Units rad/s Units rad/s 2 Units rad/s 2 Unitsrad/s 2

Explanation / Answer

here,

theta = 1.73 * t - 3.37 * t^2 + 2.31 * t^3

differentiating the equation for angular velocity

w = 1.73 - 6.74 * t + 6.93 * t^2

differentiating the equation for angular accelration

alpha = - 6.74 + 13.86 * t

a)

at t = 3.19 s

the angular velocity , w(3.19) = 1.73 - 6.74 * 3.19 + 6.93 * 3.19^2 = 50.7 rad/s

b)

at t = 9.91 s

the angular velocity , w(9.91) = 1.73 - 6.74 * 9.91 + 6.93 * 9.91^2 = 615.5 rad/s

c)

the average accelration , alpha = ( w(9.91) - w(3.19))/(9.91 - 3.19) rad/s^2

alpha = 84 rad/s^2

d)

at t = 3.19 s

the angular accelration , alpha(3.19) = - 6.74 + 13.86 * 3.19 rad/s

alpha(3.19) = 37.5 rad/s^2

e)

at t = 9.91 s

the angular accelration , alpha(9.91) = - 6.74 + 13.86 * 9.91 rad/s

alpha(9.91) = 130.6 rad/s^2

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