What type of machines are you talking about?
Answer:
α =
Explanation:
Applying the equations of motion to determine angular acceleration of the unit,
The sum of moments about O is equal to the product of angular acceleration and moment of inertia
∑Mo = Io*α
Taking the anticlockwise direction as positive moment,
= ( -(1150) + (1400) ) * (0.5 / 2) + ( (475) - (650) ) * (0.3 / 2) - F = Io*α
= 36.5 - (2.5 N.m) =*α
NOTE: moment of inertia of the pulleys in this instance =
Hence, 33.75 = * α
Solving, α =
Answer:
Δt'/ T% = 90.3%
Explanation:
Simple harmonic movement is described by the expression
x = A cos (wt)
we find the time for the two points of motion
x = - 0.3 A
-0.3 A = A cos (w t₁)
w t₁ = cos -1 (-0.3)
remember that angles are in radians
w t₁ = 1.875 rad
x = 0.3 A
0.3 A = A cos w t₂
w t₂ = cos -1 (0.3)
w t₂ = 1,266 rad
Now let's calculate the time of a complete period
x= -A
w t₃ = cos⁻¹ (-1)
w t₃ = π rad
this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period
T = 2 t₃
T = 2π / w s
now we can calculate the fraction of time in the given time interval
Δt / T = (t₁ -t₂) / T
Δt / T = (1,875 - 1,266) / 2pi
Δt / T = 0.0969
This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is
Δt'/ T = 1 - 0.0969
Δt '/ T = 0.903
Δt'/ T% = 90.3%
5.438 to the nearest tenth is 5.4