Question

During a certain time interval, the angular position of a swinging door is described by θ = 4.96 + 10.7t + 1.93t², where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times.
(a) t = 0

θ	= rad
ω	= rad/s
α	= rad/s2

(b) t = 2.99 s

θ	=    rad
ω	=    rad/s
α	=    rad/s2

Answer 1

Given,

theta = 4.96 + 10.7t + 1.93t^2

a)t = 0 ; putting t = 0 in given eqn we get

theta = 4.96 rad

w = d(theta)/dt

w = d(4.96 + 10.7t + 1.93t^2)/dt

w = 0 + 10.7 + 3.86 t

for t = 0

w = 10.7 rad/s

alpha = d(w)/dt = d(10.7 + 3.86 t) = 3.86

alpha = 3.86 rad/s^2

Hence, at t = 0 ; theta = 4.96 rad ; w = 10.7 rad ; alpha = 3.86 rad/s^2

b)at t = 2.99

theta = 4.96 + 10.7 x 2.99 + 1.93 x 2.99^2 = 54.21 rad

w = 10.7 + 3.86 x 2.99 = 22.01 rad/s

alpha = 3.86 rad/s^2

Hence, at t = 2.99 s ; theta = 54.21 rad ; w = 22.01 rad/s ; alpha = 3.86 rad/s^2