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

Answer:

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

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

$T_{net}=I\times\alpha$

$T_{A}r-T_{B}r=I\times \dfrac{a}{r}$

$T_{A}-T_{B}=I\times\dfrac{a}{r^2}$

$m_{A}g-m_{A}a-(m_{B}g+m_{B}a)=I\times\dfrac{a}{r^2}$

$g(m_{A}-m_{B})-a(m_{A}+m_{B})=I\times\dfrac{a}{r^2}$

$g(m_{A}-m_{B})=I\times\dfrac{a}{r^2}+a(m_{A}+m_{B})$

$g(m_{A}-m_{B})=a(\dfrac{I}{r^2}+(m_{A}+m_{B}))$

$a=\dfrac{g(m_{A}-m_{B})}{(\dfrac{I}{r^2}+(m_{A}+m_{B}))}$

Put the value into the formula

$a=\dfrac{9.8\times(6.00-3.00)}{\dfrac{0.220}{(0.120)^2}+(6.00+3.00)}$

$a=1.21\ m/s^2$

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