Question

illustrates an Atwood's machine. Let the masses of blocks A and B be 6.00 kg and 3.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.220 kg⋅m2, and the radius of the wheel be 0.120 m. There is no slipping between the cord and the surface of the wheel.

Answer 1

The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.

Step-by-step explanation:

Given that,

Mass of block A = 6.00 kg

Mass of block B = 3.00 kg

Moment of inertia = 0.220 kg.m²

Radius = 0.120 m

Suppose we need to find the the magnitude of the linear acceleration of block A

Let a is the acceleration of the blocks.

Let
`T_(a)` and
`T_(b)` are the tension in the A and B cord.

According to figure,

We need to calculate the magnitude of the linear acceleration of block A

Net force acting on block A,

`F_(A)=m_(A)g-T_(A)`

`m_(A) a=m_(A)g-T_(A)`

`T_(A)=m_(A)g-m_(A)a`...(I)

Net force acting on block B,

`F_(B)=T_(B)-m_(B)g`

`m_(B)a=T_(B)-m_(B)g`

`T_(B)=m_(B)a+m_(B)g`...(II)

Net torque acting on pulley

`T_(net)=I*\alpha`

`T_(A)r-T_(B)r=I* (a)/(r)`

`T_(A)-T_(B)=I*(a)/(r^2)`

`m_(A)g-m_(A)a-(m_(B)g+m_(B)a)=I*(a)/(r^2)`

`g(m_(A)-m_(B))-a(m_(A)+m_(B))=I*(a)/(r^2)`

`g(m_(A)-m_(B))=I*(a)/(r^2)+a(m_(A)+m_(B))`

`g(m_(A)-m_(B))=a((I)/(r^2)+(m_(A)+m_(B)))`

`a=(g(m_(A)-m_(B)))/(((I)/(r^2)+(m_(A)+m_(B))))`

Put the value into the formula

`a=(9.8*(6.00-3.00))/((0.220)/((0.120)^2)+(6.00+3.00))`

`a=1.21\ m/s^2`

Hence, The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.