91Ó°ÊÓ

The angle made by the ramp,$\mathrm{θ}=30°$

Mass of object,$m=0.5\text{kg}$

Radius, of the object,$r=6.00\text{cm}=0.06\text{m}$

As we are given the graph of velocity vs time, determine the acceleration from the graph. Using the formula for the acceleration in terms of moment of inertia, find the required answer.

The linear acceleration, is given as-

$$\begin{gathered} a_{linear}=-\frac{g×\sin \mathrm{θ}}{1+\left(\frac{I}{m{r}^2}\right)} \\ a=\frac{\mathrm{Δ}v}{\mathrm{Δ}t} \end{gathered}$$

From the graph,

$$\begin{gathered} {\text{v}}_{\text{i}}=0,{\text{t}}_{\text{i}}=0 \\ {\text{v}}_{\text{f}}=3.5\frac{\text{m}}{\text{s}},{\text{t}}_{\text{f}}=1 \\ \text{a}=\frac{\mathrm{Δ}\text{v}}{\mathrm{Δ}\text{t}} \\ a=\frac{{\displaystyle v_f-v_i}}{{\displaystyle t_f-t_i}} \\ a=\frac{{\displaystyle 0-3.5}}{{\displaystyle 1\text{s}-0}} \\ a=-3.5m/s^2 \end{gathered}$$

As we know, the linear acceleration is given as-

$a_{linear}=-\frac{g×\sin \mathrm{θ}}{1+\left(\frac{I}{m{r}^2}\right)}$

For the given values, the equation becomes-

$$\begin{gathered} -3.5m/s^2=-\frac{9.8m/s^2×\sin 30}{1+\left(\frac{\text{I}}{0.5\text{kg}×{\left(0.06\text{m}\right)}^2}\right)} \\ 3.5m/s^2=\frac{4.9m/s^2}{1+\left(\frac{\text{I}}{1.8×{10}^{-3}{\text{kgm}}^2}\right)} \\ 3.5m/s^2+\left(\frac{3.5m/s^2}{1.8×{10}^{-3}{\text{kgm}}^2}\right)\text{I}=4.9m/s^2 \end{gathered}$$

$7.2×{10}^{-4}{\text{kgm}}^2$

So, the rotational inertia of the given object is $7.2×{10}^{-4}{\text{kgm}}^2$.