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A 2.10 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 3.20 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.900 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? Can you explain why the tensions are not the same even though it is the same cord.

Respuesta :

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 .

okpalawalter8

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

What is the tension in each part of the cord and the moment of inertia of the pulley?

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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