Respuesta :
Answer:
Explanation:
First of all we shall calculate the acceleration of the system . Let it be a .
s = ut + 1/2 a t²
1.2 = 1/2 x a x .9²
a = 2.963
A ) Applying 2nd law of motion on textbook
T₁ = ma , m is mass of the textbook a is its acceleration
T₁ = 1.9 x 2.963
= 5.63 N .
B ) Applying 2nd law of motion on hanging book
mg - T₂ = ma , m is mass of hanging book , a is its acceleration
T₂ = mg - ma
= m ( g - a )
= 3.1 x ( 9.8 - 2.963 )
= 21.2 N
C )
Torque on the pulley
= ( T₂ - T₁ ) x R
R is radius of the pulley
= (21.2 - 5.63 ) x .075
= 1.16775 Nm
angular acceleration of pulley
α = linear acceleration / radius
= 2.963 / .075
= 39.5 rad / s²
torque on pulley = moment of inertia x angular acceleration
1.16775 = moment of inertia x 39.5
moment of inertia = 29.56 x 10⁻³ kg m² .
(a) The tension of the chord attached to the textbook is 5.624 N.
(b) The tension of the chord attached to the hanging book is 21.2 N.
(c) The moment of inertia of the pulley about its rotation axis is 0.0296 kgm².
The given parameters:
- Mass of the text book, m = 1.90 kg
- Diameter of the pulley, d = 0.15 m
- Mass of hanging book = 3.10 kg
- distance traveled by the book, s = 1.2 m
- time of motion of the book, t = 0.9 s
The acceleration of the book at the given displacement and time is calculated as;
[tex]s = ut \ + \ \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2} \\\\a = \frac{2 \times 1.2 }{0.9^2} \\\\a = 2.96 \ m/s^2[/tex]
The tension of the chord attached to the textbook is calculated as follows;
T = ma
T = 1.9 x 2.96
T = 5.624 N
The tension of the chord attached to the hanging book is calculated as follows;
T = mg - ma
T = m(g - a)
T = 3.1(9.8 - 2.96)
T = 21.2 N
The torque on the pulley is calculated as;
[tex]\tau = \Delta T R\\\\ \tau = (T_2-T_1)R\\\\ \tau = (21.2 - 5.624) \times 0.075\\\\ \tau = 1.168 \ Nm[/tex]
The angular acceleration of the books is calculated as;
[tex]\alpha = \frac{a}{R} \\\\\alpha = \frac{2.964}{0.075}\\\\\alpha =39.52 \ rad/s^2[/tex]
The moment of inertia of the pulley about its rotation axis is calculated as follows;
[tex]\tau = I \alpha\\\\I = \frac{\tau }{\alpha} \\\\I = \frac{1.168}{39.52} \\\\I = 0.0296 \ kg m^2[/tex]
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