16. The angular position of a swinging door is deseribed by 0(t) = 5+ 10t + 2t², where 0
is in radians and t is in seconds. Determine the angular position, angular speed, and angular
acceleration of the door (a) at =0 and (b) at t = 3s.
To find the angular position at t=0, we substitute t=0 into the equation:
θ(t=0) = 5 + 10(0) + 2(0)^2
θ(t=0) = 5 + 0 + 0
θ(t=0) = 5
Therefore, the angular position at t=0 is 5 radians.
To find the angular speed at t=0, we take the derivative of the equation with respect to time:
ω(t) = dθ/dt = 10 + 4t
Substituting t=0 into the angular speed equation:
ω(t=0) = 10 + 4(0)
ω(t=0) = 10 + 0
ω(t=0) = 10
Therefore, the angular speed at t=0 is 10 radians per second.
To find the angular acceleration at t=0, we take the second derivative of the equation with respect to time:
α(t) = d²θ/dt² = d/dt(10 + 4t) = 4
Therefore, the angular acceleration at t=0 is 4 radians per second squared.
To find the angular position at t=3s, we substitute t=3 into the equation:
θ(t=3) = 5 + 10(3) + 2(3)^2
θ(t=3) = 5 + 30 + 18
θ(t=3) = 53
Therefore, the angular position at t=3s is 53 radians.
To find the angular speed at t=3s, we substitute t=3 into the angular speed equation:
ω(t=3) = 10 + 4(3)
ω(t=3) = 10 + 12
ω(t=3) = 22
Therefore, the angular speed at t=3s is 22 radians per second.
To find the angular acceleration at t=3s, we substitute t=3 into the angular acceleration equation:
α(t=3) = 4
Therefore, the angular acceleration at t=3s is 4 radians per second squared.