Figure $\mathbf{10}\mathbf{-}\mathbf{46}$ shows particles $\mathbf{1}$ and $\mathbf{2}$, each of mass m, fixed to the ends of a rigid massless rod of length ${\mathbf{L}}_{\mathbf{1}}\mathbf{+}{\mathbf{L}}_{\mathbf{2}}$ , with ${\mathbf{L}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{\text{\hspace{0.17em}cm}}$ and${\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{80}\mathbf{\text{\hspace{0.17em}cm}}$ . The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of

(a) particle$\mathbf{\text{1}}$ and

(b) particle$\mathbf{\text{2}}$ ?

1. The gravitational acceleration is,$g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.
2. The length of particle 1 is,${\mathrm{L}}_1=0.20\text{\hspace{0.17em}m}$.
3. The length of particle 2 is, ${\mathrm{L}}_2=0.80\text{\hspace{0.17em}m}$.

We can use the formula for the torque in terms of weight and distance as well as formula in terms of rotational inertia and angular acceleration to find the acceleration of each particle.

$\begin{array}{c}\mathbf{\tau }\mathbf{=}\mathbf{mgL}\\ \mathbf{=}\mathbf{I\alpha }\mathbf{}\end{array}$

Moment of inertia,$\mathbf{I}\mathbf{=}\mathbf{}{\mathbf{mL}}^{\mathbf{2}}$

To find the initial acceleration of particles,

Let the direction of rigid support move in counter clockwise positive direction. Also, masses of both the particles are the same. Therefore, the angular acceleration $\alpha$satisfies the condition as follows,

$\tau =mg{L}_1-mg{L}_2$

Also, we know,$\mathrm{\tau }=\mathrm{I\alpha }$

And Moment of inertia,$\mathrm{I}=\mathrm{m}({\mathrm{L}}_1^2+{\mathrm{L}}_2^2)$

Therefore, putting values you get,

$\begin{array}{c}{\mathrm{mgL}}_1-{\mathrm{mgL}}_2=\mathrm{I\alpha }\\ \mathrm{\alpha }=\frac{({\mathrm{L}}_1-{\mathrm{L}}_2)\mathrm{mg}}{\mathrm{m}({\mathrm{L}}_1^2+{\mathrm{L}}_2^2)}\\ =\frac{({\mathrm{L}}_1-{\mathrm{L}}_2)\mathrm{g}}{({\mathrm{L}}_1^2+{\mathrm{L}}_2^2)}\\ \mathrm{\alpha }=\frac{\mathrm{g}({\mathrm{L}}_1-{\mathrm{L}}_2)}{{{\mathrm{L}}_1}^2+{{\mathrm{L}}_2}^2}\end{array}$

Substitute all the value in the above equation,

$\begin{array}{c}\alpha =\frac{(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})(0.20\text{\hspace{0.17em}m}-0.80\text{\hspace{0.17em}m})}{{\left(0.20\text{\hspace{0.17em}m}\right)}^2+{\left(0.80\text{\hspace{0.17em}m}\right)}^2}\\ =-8.65\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Negative sign indicates that the system will start moving in clockwise direction. As it is measured at time zero, their instantaneous velocity will be also zero.

Therefore, Acceleration at point 1 and 2 can be calculated as

$\left|{\mathrm{a}}_1\right|=\left|\mathrm{\alpha }\right|{\mathrm{L}}_1$

i.e.,

$\begin{array}{c}\left|{\mathrm{a}}_1\right|=(-8.65\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}})\left(0.20\text{\hspace{0.17em}m}\right)\\ =1.7\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

And
$\left|{\text{a}}_2\right|=\left|\mathrm{\alpha }\right|{\text{L}}_2$

i.e.,

$\begin{array}{c}\left|{\mathrm{a}}_1\right|=(-8.65\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}})\left(0.80\text{\hspace{0.17em}m}\right)\\ =\mathrm{1.76.9}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Hence, theInitial acceleration of particle 1 is,${\text{a}}_1=1.7\text{\hspace{0.17em}m}/{\text{s}}^2$.

And Initial acceleration of particle 2 is, ${\text{a}}_2=6.9\text{\hspace{0.17em}m}/{\text{s}}^2$.