A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter
is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?
1 answer:
Given that,
Mass of textbook Mb = 2kg
Initially at rest Ub=0m/s
diameter of pulley d = 0.15m
Then, the radius is r = d/2 =0.15/2 = 0.075m
A hang book of mass Mh = 3kg
Initial at rest too Uh = 0m/s
Distance moved = 1.2m in t = 0.8s
Check attachment for free body diagram and better understanding
Analysis of the mass of textbook on table
Using Newton second law
ΣFx = m•ax
T1 = Mb•ax
T1 = 2•ax, ............ equation 1
Now, analysis of the hanged body
Using the same newton's law
ΣFy = m•ay
T2 — Mh•g = —3•ay,
ay is negative because it is moving in the negative direction
Then, T2 = Mh•g — 3•ay
T2 = 3×9.81 —3•ay
T2 = 29.43 — 3•ay, ............... equation 2
Now, the body moves 1.2m in t=0.8second
The initial velocity of the body is 0m/s
Using equation of motion to calculate the acceleration (a)
S=ut+½at²
1.2 = 0•t + ½ × a × 0.8²
1.2 = 0 + 0.32a
0.32a =1.2
a = 1.2/0.32
a = 3.75m/s²
Since the body are connected by an I inextensive string, their acceleration are the same i.e. ax=ay=a=3.75m/s²
So, back to equation 1
T1 = 2•ax
T1 = 7.5 N
Also, back to equation 2
T2 = 29.43 — 3•ay
T2 = 29.43 — 3 × 3.75
T2 = 29.43 — 11.25
T2 = 18.18 N
b. Torque?
Modelling the pulley as a rigid body
Then, applying equilibrium of torque
Clockwise torque = anti-clockwise torque
Σ τ = (T2-T1)r = Iα
Where α is angular acceleration
Relationship between angular acceleration and radial acceleration is given as a=αr
Therefore,
(T2-T1)r = Iα
Since α=a/r
(T2-T1) = Ia/r
Cross multiply
(T2-T1)r² = Ia
(18.18 — 7.5) × 0.075² = I(3.75)
10.68×0.075² = 3.75I
0.060075 = 3.75I
Then, I = 0.060075/3.75
I = 0.01602 kgm²
I ≈ 0.016 kgm²
Then, the moment of inertia is 0.016 kgm²
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Answer:
a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.
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part a)
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so now we can substitute the corresponding values:
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but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:
θ=270°-60°=210°
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and the time up stream:
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so
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