Question

Question: Figure gives the speed vversus time tfor 0.500 kg a object of radius 6.00 cmthat rolls smoothly down a 30 ⁰ramp. The scale on the velocity axis is set by v_s = 4.0 m/sWhat is the rotational inertia of the object?

Short answer

Answer

Rotational inertia of the object is$7.2\times {10}^{-4}{\text{kgm}}^2$

Step-by-step solution

Given

The angle made by the ramp,$\theta =30°$

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

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

To understand the concept

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-

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

Calculate the acceleration from the graph

From the graph,

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

Calculate the rotational inertia of the given object

As we know, the linear acceleration is given as-

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

For the given values, the equation becomes-

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

$7.2\times {10}^{-4}{\text{kgm}}^2$

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