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erinwebsterrr
26.06.2020 •
Physics
g 1.90 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 3.10 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.900 s . Part A What is the tension in the part of the cord attached to the textbook? Express your answer with the appropriate units. T1 = nothing nothing Request Answer Part B What is the tension in the part of the cord attached to the hanging book? Express your answer with the appropriate units. T2 = nothing nothing Request Answer Part C What is the moment of inertia of the pulley about its rotation axis? Express your answer with the appropriate units. I = nothing nothing Request Answer Provide Feedback
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Ответ:
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 .
Ответ: