The Atwood’s machine Figure P10.59 Illustrates an Atwood’s machine. Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks A and B be 4.00 kg and 2.00 kg, respectively, the moment of inertia of the wheel about its axis be $\mathbf{0}\mathbf{.}\mathbf{220}\mathbf{kg}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}$, and the radius of the wheel be 0.120 m.

The angular acceleration of any moving body can be described as the changing rate of angular velocity of the body with respect to the time. The unit of angular acceleration is radians per square second.

The angular acceleration of the gritstone will be by the formula of motion,

$\begin{array}{r}{\omega }_{\mathrm{z}}={\omega }_{0\mathrm{z}}+{\alpha }_{\mathrm{z}}t\\ \sum {\tau }_{\mathrm{z}}=I{\alpha }_{\mathrm{z}}\\ f_k={\mu }_{k}n\\ {\mu }_k=-\frac{MR{\alpha }_{\mathrm{z}}}{2n}\end{array}$

Where, n is force ,${\omega }_z$ final rotation speed,${\omega }_{0z}$ starting rotation speed,${\alpha }_z$ rotation acceleration,${\tau }_z$ torque,${\mu }_k$ coefficient of friction,$f_k$ friction force, I moment of inertia, and t time. are mass M and radius R of the object.

Moment of inertia =$0.220\mathrm{kg}.{\mathrm{m}}^2$

Block A mass =$4.00\mathrm{kg}$

Block B mass = 2.00 kg

Radius of the wheel = 0.120 m

For A, the force will be

$\begin{array}{r}\sum F_y=m{a}_y\\ m_{A}g-T_A=m_{A}a\end{array}$

For B, the force will be

$\begin{array}{r}\sum F_y=m{a}_y\\ T_B-m_{B}g=m_{B}a\end{array}$

For wheel, the torque will be

$\begin{array}{r}\sum {\tau }_{\mathrm{z}}=I{\alpha }_{\mathrm{z}}\\ T_{A}R-T_{B}R=I{\alpha }_{\mathrm{z}}\\ T_{A}R-T_{B}R=I(\mathrm{a}/R)\\ T_A-T_B=\left(\frac{I}{R^2}\right)a\end{array}$

Here,

$F_y$is the force.

$T_A$is the tension on block A,

$T_B$is the tension on block,

$m_A$is the mass of block A,

$m_B$is the mass of block B,

g is the gravity,

a is the accelaration,

R is the radius of wheel,

${\alpha }_z$is the rotational accelration,

${\tau }_z$is the torque

From the above three equation we will get,

$\begin{array}{r}\left(m_A-m_B\right)g=\left(m_A+m_B+\frac{I}{R^2}\right)a\\ a=\frac{\left(m_A-m_B\right)g}{\left(m_A+m_B+I/R^2\right)}\\ a=\left(\frac{4.00\mathrm{kg}-2.00\mathrm{kg}}{4.00\mathrm{kg}+2.00\mathrm{kg}+\left(\frac{0.220\mathrm{kg}\cdot {\mathrm{m}}^2}{(0.120\mathrm{m}{)}^2}\right)}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ a=0.921\mathrm{m}/{\mathrm{s}}^2\end{array}$

Further, the angular acceleration of the wheel C will be,

$\begin{array}{r}\alpha =\frac{0.921\mathrm{m}/{\mathrm{s}}^2}{0.120\mathrm{m}}\\ \alpha =7.68\mathrm{rad}/{\mathrm{s}}^2\end{array}$

Linear acceleration of blocks A and B will be,

$\begin{array}{r}{T}_A=m_A(g-a)\\ T_A=4.00\mathrm{kg}\left(9.80\mathrm{m}/{\mathrm{s}}^2-0.921\mathrm{m}/{\mathrm{s}}^2\right)\\ T_A=35.51\mathrm{N}\end{array}$

$\begin{array}{r}{T}_B=m_B(g+a)\\ T_B=2.00\mathrm{kg}\left(9.80\mathrm{m}/{\mathrm{s}}^2+0.921\mathrm{m}/{\mathrm{s}}^2\right)\\ T_B=21.44\mathrm{N}\end{array}$

Hence the Linear acceleration of blocks A and B are${\mathbf{T}}_{\mathbf{A}}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{51}\mathbf{N}$and${\mathbf{T}}_{\mathbf{B}}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{44}\mathbf{}\mathbf{N}$ respectively. The angular acceleration of the wheel C is$\mathbf{\alpha }\mathbf{7}\mathbf{.}\mathbf{68}\mathbf{}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ .

The direction of the tension should be different so that they can produce the torque that can able to accelerate the wheel when the block A and B accelerate.