91Ó°ÊÓ

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,

$\begin{array}{r}{\mathrm{Ó\neg }}_{\mathrm{z}}={\mathrm{Ó\neg }}_{0\mathrm{z}}+{\mathrm{Î\pm }}_{\mathrm{z}}t\\ ∑{\mathrm{Ï„}}_{\mathrm{z}}=I{\mathrm{Î\pm }}_{\mathrm{z}}\\ f_k={\mathrm{μ}}_{k}n\\ {\mathrm{μ}}_k=−\frac{MR{\mathrm{Î\pm }}_{\mathrm{z}}}{2n}\end{array}$

Where, n is force ,${\mathrm{Ó\neg }}_z$ final rotation speed,${\mathrm{Ó\neg }}_{0z}$ starting rotation speed,${\mathrm{Î\pm }}_z$ rotation acceleration,${\mathrm{Ï„}}_z$ torque,${\mathrm{μ}}_k$ 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

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

For B, the force will be

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

For wheel, the torque will be

$\begin{array}{r}∑{\mathrm{Ï„}}_{\mathrm{z}}=I{\mathrm{Î\pm }}_{\mathrm{z}}\\ T_{A}R−T_{B}R=I{\mathrm{Î\pm }}_{\mathrm{z}}\\ T_{A}R−T_{B}R=I(\mathrm{a}/R)\\ T_A−T_B=\left(\frac{I}{R^2}\right)a\end{array}$

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,

${\mathrm{Î\pm }}_z$is the rotational accelration,

${\mathrm{Ï„}}_z$is the torque

From the above three equation we will get,

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

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

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

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$\mathbf{Î\pm }\mathbf{7}\mathbf{.}\mathbf{68}\mathbf{}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ .

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.