Question

How to do this one? please help me!

The angular position of a point on the rim of a rotating wheel is given by theta = 3.31t - 6.48t^2 + 1.20t^3, where theta is in radians and t is in seconds. What are the angular velocities at (a) t = 3.56 s and (b) t = 9.39 s? (c) What is the average angular acceleration for the time interval that begins at t = 3.56 s and ends at t = 9.39 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units

Answer 1

Given :

$\theta$ = 3.31 t - 6.48 t² + 1.2 t³

taking derivative both side relative to "t"

d$\theta$/dt = 3.31 - 12.96 t + 3.6 t²

W = 3.31 - 12.96 t + 3.6 t²                            eq-1

a) at t = 3.56

W = 3.31 - 12.96 (3.56) + 3.6 (3.56)²

W = 2.797 rad/s

b)

at t = 9.39

W = 3.31 - 12.96 (9.39) + 3.6 (9.39)²

W = 199.04 rad/s

C)

Change in angular velocity = 199.04 - 2.797 = 196.24 rad/s

time interval = t = 9.39 - 3.56 = 5.83

average angular acceleration = change in angular velocity / time interval = 196.24 / 5.83 = 33.7 rad/s

d)

taking derivative of eq-1 relative to "t"

dW/dt = 0 - 12.96 + 7.2 t

$\alpha$ = 7.2 t - 12.96                         eq-2

at t = 3.56

$\alpha$ = 7.2 (3.56) - 12.96    

    $\alpha$ = 12.67 rad/s²

e)

at t = 9.39

$\alpha$ = 7.2 (9.39) - 12.96    

$\alpha$ = 54.65 rad/s²