3. When the procedure is repeated with a third line how will it distinguish whether the location of the center of gravity is acc
1 answer:
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Answer:
Explanation:
Given that,
When Mass of block is 12kg
M = 12kg
Block falls 3m in 4.6 seconds
When the mass of block is 24kg
M = 24kg
Block falls 3m in 3.1 seconds
The radius of the wheel is 600mm
R = 600mm = 0.6m
We want to find the moment of inertia of the flywheel
Taking moment about point G.
Then,
Clockwise moment = Anticlockwise moment
ΣM_G = Σ(M_G)_eff
M•g•R - Mf = I•α + M•a•R
Relationship between angular acceleration and linear acceleration
a = αR
α = a / R
M•g•R - Mf = I•a / R + M•a•R
Case 1, when y = 3 t = 4.6s
M = 12kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 4.6²
3 × 2 = 4.6²a
a = 6 / 4.6²
a = 0.284 m/s²
M•g•R - Mf = I•a / R + M•a•R
12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6
70.632 - Mf = 0.4726•I + 2.0448
Re arrange
0.4726•I + Mf = 70.632-2.0448
0.4726•I + Mf = 68.5832 equation 1
Second case
Case 2, when y = 3 t = 3.1s
M= 24kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 3.1²
3 × 2 = 3.1²a
a = 6 / 3.1²
a = 0.6243 m/s²
M•g•R - Mf = I•a / R + M•a•R
24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6
141.264 - Mf = 1.0406•I + 8.99
Re arrange
1.0406•I + Mf = 141.264 - 8.99
1.0406•I + Mf = 132.274 equation 2
Solving equation 1 and 2 simultaneously
Subtract equation 1 from 2,
Then, we have
1.0406•I - 0.4726•I = 132.274 - 68.5832
0.568•I = 63.6908
I = 63.6908 / 0.568
I = 112.13 kgm²
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s