Answer:
Explanation:
acceleration of the system
s = 1/2 a t²
a =2s /t²
= 2x 1.2 / .9²
= 2.96 m /s²
Let the tension be T₁ and T₂ in the cord
For movement of 3.2 kg book
3.2 g - T₁ = 3.2 a
For movement of 2.1 kg book
T₂ - 2.1 g = 2.1 a
Adding
(T₂ - T₁ ) + 1.1 = 5.3 a
For rotation of pulley
(T₂ - T₁ ) R = I a /R ( I is moment of inertia of pulley )
(T₂ - T₁ )= I a /R²
= I x 2.96 /.075²
Putting this value in earlier equation
I x 2.96 /.075² + 1.1 = 5.3 a
526.22 I + 1.1 = 15.7
526.22 I = 14.6
I = 2.77 X 10⁻² kg m²
Tension are not the same because of friction between pulley and cord which rotates the pulley .
The tension in each part of the cord and the moment of inertia of the pulley is mathematically given as
T2 - T1 )= 14.576N
I = 2.77 X 10^{-2} kg m^2
Question Parameter(s):
A 2.10 kg textbook rests on a frictionless
A cord attached to the book passes over a pulley whose diameter is 0.150 m,
the books are observed to move 1.20 m in 0.900 s.
Generally, the equation for the acceleration is mathematically given as
a =2s /t^2
Therefore
a= 2x 1.2 / .9^2
a= 2.96 m /s^2
Where tension is
(T2 - T1 ) + 1.1 = 5.3 a
(T2 - T1 )= I a /R^2
(T2 - T1 )= I x 2.96 /.075^2
Therefore
I x 2.96 /.075^2 + 1.1 = 5.3 a
I = 2.77 X 10^{-2} kg m^2
In conclusion
(T2 - T1 )= (2.77 * 10^{-2)) * (2.96 /0.075^2)
T2 - T1 )= 14.576N
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