In Fig. 10-50, two$\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{\text{kg}}$blocks are connected by a mass-less string over a pulley of radius$\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{\text{cm}}$and rotational inertia.$\mathbf{7}\mathbf{.40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\text{kg/m}}}^{\mathbf{2}}$The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley’s axis is frictionless. When this system is released from rest, the pulley turns through$\mathbf{0}\mathbf{.130}\mathbf{\text{rad}}$in$\mathbf{\text{91.0ms}}$and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley’s angular acceleration, (b) the magnitude of either block’s acceleration, (c) string tension${\mathit{T}}_{\mathbf{1}}$, and (d)${\mathit{T}}_{\mathbf{2}}$string tension?

Mass of block$\text{m}=6.20\text{kg}$

Radius of pulley$\text{R}=2.40\text{cm}=2.40\times {10}^{-2}\text{m}$

Rotational inertia$\text{I}=7.40\times {10}^{-4}{\text{kg.m}}^2$

Angular displacement$\theta =0.130\text{\hspace{0.17em}rad}$

Time$\text{t}=91.0\text{ms}=91.0\times {10}^{-3}\text{s}$

Using the second kinematic equation of rotational motion, we can find the angular acceleration of the pulley. Using this angular acceleration into the relation between linear acceleration and angular acceleration, we can find the magnitude of each block’s acceleration. We apply Newton’s second law of rotation to write the equation of net force exerted on the hanging block. Using this equation, we can find the string tension using ${\text{T}}_{\text{1}}\text{.}$net torque equation for the pulley; we can find the string tension${\text{T}}_{\text{2}}$.

Formula:

$\theta ={\omega }_0\text{t}+\frac{1}{2}\alpha {\text{t}}^2$

$a=\alpha \text{R}$

We have second kinetic equation in rotational motion:

$\theta ={\omega }_0\text{t}+\frac{1}{2}\alpha {\text{t}}^2$

$\Rightarrow \left(0.130\text{rad}\right)=0(91.0\times {10}^{-3}\text{s})+\frac{1}{2}\alpha {(91.0\times {10}^{-3}\text{s})}^2$

$\Rightarrow (0.130\text{rad})=\frac{1}{2}\alpha {(91.0\times {10}^{-3}\text{s})}^2$

$\Rightarrow \alpha =\frac{2(0.130\text{rad})}{{(91.0\times {10}^{-3}\text{s})}^2}$

$\Rightarrow \alpha =31.4\frac{\text{rad}}{{\text{s}}^{\text{2}}}$

Therefore, the magnitude the pulley’s acceleration is$31.4\text{\hspace{0.17em}rad}/{\text{s}}^2$

As the pulley is frictionless, both blocks have the same acceleration. Therefore,

$\text{a}=\alpha \text{R}$

$\text{a}=\left(31.4\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)(2.40\times {10}^{-2}\text{m})$

$\text{a}=0.754\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Therefore, the magnitude of either block’s acceleration is$0.754\text{m}/{\text{s}}^2$ .

Using Newton’s second law of rotation for hanging block, we have

$\text{mg}-{\text{T}}_1=\text{ma}$

$\Rightarrow {\text{T}}_1=\text{mg}-\text{ma}$

$\Rightarrow {\text{T}}_{\text{1}}=\text{m}(\text{g}-\text{a})$

$\Rightarrow {\text{T}}_{\text{1}}=(6.20\text{\hspace{0.17em}kg})(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}-0.754\frac{\text{m}}{{\text{s}}^{\text{2}}})$

$\Rightarrow {\text{T}}_{\text{1}}=56.1\text{\hspace{0.17em}N}$

Therefore, string tension${\text{T}}_{\text{1}}$ is$56.1\text{N}$

For the pulley, net torque equation is

${\text{RT}}_1-{\text{RT}}_2=\text{I}\alpha$

$\Rightarrow \text{R}({\text{T}}_1-{\text{T}}_2)=\text{I}\alpha$

$\Rightarrow (T_1-T_2)=\frac{\text{I}\alpha }{\text{R}}$

$\Rightarrow {\text{T}}_2={\text{T}}_1-\frac{\text{I}\alpha }{\text{R}}$

$\Rightarrow {\text{T}}_2=(56.1\text{N})-\frac{(7.40\times {10}^{-4}{\text{kg.m}}^2)\left(31.4\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)}{(2.40\times {10}^{-2}\text{m})}$

$\Rightarrow {\text{T}}_{\text{2}}=55.1\text{\hspace{0.17em}N}$

Therefore, string tension${\text{T}}_{\text{2}}$ is$55.1\text{\hspace{0.17em}N}.$