A general contractor has received plans for a new high-rise hotel in an urban area. The hotel will be 12 stories tall and will h
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Answer:
1) 1.71 rad/s
2) -6.22 rad/s²
Explanation:
Choose point C to be the origin.
Using geometry, we can show that the coordinates of point A are:
(a cos 30°, a sin 30° − b)
Therefore, the coordinates of point D at time t are:
(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))
The angle formed by CB with the x-axis is therefore:
tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))
1) Taking the derivative with respect to time, we can find the angular velocity:
sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.
sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²
sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²
dθ/dt = (200) (6) (1/2) / 350
dθ/dt = 600 / 350
dθ/dt = 1.71 rad/s
2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴
At t = 0:
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)
d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)
Plugging in values:
d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° (1.71) = -4.24
d²θ/dt² = -6.22 rad/s²
Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
