In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
pped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.
1 answer:
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
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Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Answer:
This can be the FBD of the bag.
(my bag looks more like a box tho ^^")
