Answer:
(A) 4.75 m/s^{2}
(b) 81.5°
Explanation:
radius (r) = 0.38 m
initial angular velocity (ω') = 1.5 rad/s
angular acceleration (α) = 2 rad/s^{2}
time (t) = 0.5 s
(A) tangential acceleration (a) = α x r
a = 2 x 0.38 = 0.76 rad/s^{2}
- radial acceleration (a') =
final angular velocity (ω) = ω + αt
ω = 1.5 + (2 x 0.5) = 3.5 m/s
a' = = 4.7 m/s^{2}
- total acceleration =
total acceleration = = 4.75 m/s^{2}
(B) tanθ
θ=
θ= = 81.5°
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
Answer:
Explanation:
Given data
Length L=2.5 m
Radius R=d/2=30/2 = 15 mm
Torque based on allowable stress
Allowable shear stress τ=50 Mpa
Allowable torque T=(π/2)τc³
Torque based on allowable angle of twist
Allowable Angle of twist
Ф=7.5°
Ф=7.5×(π/180)=130.90×10⁻³ rad
Allowable torque
T=(GJФ)/L
T=(G(π/2)c⁴)Ф)/L
T=(πGc⁴Ф)/2
Maximum Power Transmitted
Maximum power transmitted is given by
We are given the data for the angle of incidence and angle of refraction.
The first part of the problem is to plot the deviation and the angle of incidence. The minimum deviation is 37 with 65 degrees as the corresponding angle of incidence. The maximum deviation is 69 with a corresponding angle of incidence of 30 degrees. It is not advisable to use small values for the angle of incidence since it would result to a higher deviation from Snell's Law.