College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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In question 1 can you explain part b
I don’t understand how the moment of inertia was deduced then on the right it becomes negative and the R becomes squared please explain this
![Problem 1-Friction, pulley, and drag
TA
(a) Fr
"
B
free-body diagrams
Imsg
When the system is at rest, the pulley does not
rotate and TA=TB = T
For object A
For object B
There for
:
MA max
(b) MB/MA
=7m₂ = m₂
A
крутв
=
F = T ≤ Ffr, max = M₂MAg
T = mB9
<
m3g = us mag
0.5
us
2 (MB/MA) max
1.0
For the pulley: (TB-TA) R = Iα = Ia/R
For object A:
TA-MK mag.
= MA a
For object B.
and so: (m_MA = MB)
TA-MK mag + mag-/B = (MA+MB) 2
- Ia/R² + (1-μk) mg = 2m ²
I= _ 2 m² = (1-4₂) ma
2/R²
mR² -[(1-₂) 2 -2]
m = 1 kg; R = 0.1m; 2= 19, 1=0.4
2
I = MR² [0.6-5-2] - MR² = 1x10 " kgm²
=
(C) With drag:
dv
тв
dt
тв д-Тве тва
= mBg - T₂ - bv;
MBg-TB = M₁₂ 2
V₁ =
mg
5 b
therefore
mg
b= = = 0.2
5VT](https://content.bartleby.com/qna-images/question/6099d21a-e15a-47f8-adbb-0c871c33581f/daa6d810-fd12-4518-93f2-6677658d1098/konuhvi_processed.jpeg)
Transcribed Image Text:Problem 1-Friction, pulley, and drag
TA
(a) Fr
"
B
free-body diagrams
Imsg
When the system is at rest, the pulley does not
rotate and TA=TB = T
For object A
For object B
There for
:
MA max
(b) MB/MA
=7m₂ = m₂
A
крутв
=
F = T ≤ Ffr, max = M₂MAg
T = mB9
<
m3g = us mag
0.5
us
2 (MB/MA) max
1.0
For the pulley: (TB-TA) R = Iα = Ia/R
For object A:
TA-MK mag.
= MA a
For object B.
and so: (m_MA = MB)
TA-MK mag + mag-/B = (MA+MB) 2
- Ia/R² + (1-μk) mg = 2m ²
I= _ 2 m² = (1-4₂) ma
2/R²
mR² -[(1-₂) 2 -2]
m = 1 kg; R = 0.1m; 2= 19, 1=0.4
2
I = MR² [0.6-5-2] - MR² = 1x10 " kgm²
=
(C) With drag:
dv
тв
dt
тв д-Тве тва
= mBg - T₂ - bv;
MBg-TB = M₁₂ 2
V₁ =
mg
5 b
therefore
mg
b= = = 0.2
5VT
Transcribed Image Text:Problem 1- Friction, pulley, and drag
A system is composed of two masses connected by a massless cord going through a pulley, as shown in the figure. Object A,
with mass ma, lies on a horizontal plane, with coefficient of static friction = 0.5 and kinetic friction μ = 0.4. Object B, with
mass me=1.0 kg, is suspended in air. The pulley is basically a cylinder with radius R = 0.1 m. (a) Draw free-body diagrams for
the two masses and for the pulley. Find the maximum value of the mass ratio, (mB/MA)max, such that, for ma/mA<(MB/MA)max,
the system stays at rest. (b) Assume now that mB/MA= 2(MB/MA)max and that object B falls with initial acceleration a=g/5. Find
the value of the moment of inertia of the pulley and of the tension force TB exerted on object B. (c) For the same parameters
and values of Te obtained (b), suppose that object B is also acted upon by a drag force Fo=-bv, and that the object is allowed
to fall attached to the string until it reaches a constant terminal velocity, Vr= 9.8 m/s; find the value of the drag coefficient, b.
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