91Ó°ÊÓ

$$\begin{gathered} \mathrm{r}=0.25\mathrm{m} \\ {\mathrm{V}}_0=43\mathrm{m}/\mathrm{s} \\ \mathrm{x}=225\mathrm{m} \\ \mathrm{I}=0.155\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

Using kinematic equations, find the linear acceleration. Using the relationship between linear acceleration and angular acceleration, find angular acceleration, and using the formula for torque, find torque.

Formulae are as follow:

where,data-custom-editor="chemistry" $\mathrm{Ï„}$is torque, I is moment of inertia,data-custom-editor="chemistry" $\mathrm{Î\pm }$is an angular acceleration, r is radius, V is velocity, x is displacement and ais an acceleration.

Magnitude of linear acceleration is given by the following formula:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}^2={\mathrm{V}}_{\mathrm{i}}^2+2\mathrm{ax} \\ 0={43}^2+2×\mathrm{a}×225 \\ \mathrm{a}=-4.11\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Magnitude of acceleration is,

$\mathrm{a}=4.11\mathrm{m}/{\mathrm{s}}^2$

Hence, linear acceleration is 4.11 m/s².

Angular acceleration is given by the following formula:

$$\begin{gathered} \mathrm{Î\pm }=\frac{\mathrm{a}}{\mathrm{r}} \\ \mathrm{Î\pm }=\frac{4.11}{0.25} \\ \mathrm{Î\pm }=16.4\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Hence,angular acceleration is 16.4 rad/s².

Now, to findthetorque use the following formula:

$$\begin{gathered} \mathrm{Ï„}=\mathrm{I}×\mathrm{Î\pm } \\ \mathrm{Ï„}=0.155×16.4 \\ \mathrm{Ï„}=2.542\mathrm{N}.\mathrm{m} \\ \mathrm{Ï„}=2.55\mathrm{N}.\mathrm{m} \end{gathered}$$

Hence,the magnitude of the torque about the central axis due to friction on the wheel is 2.55 N.m.

Therefore, using kinematic equations, the linear acceleration can be found. Using the relationship between linear acceleration and angular acceleration, angular acceleration can be found and using the formula for torque, the torque can be found.