Question

A wheel of radius 0.250 m, which is moving initially at 43.0 m/s, rolls to a stop in 225 m. Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is 0.155 kg.m² about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

Short answer

1. Linear acceleration is 4.11 m/s².
2. Angular acceleration is 16.4 rad/s².
3. The magnitude of the torque about the central axis due to friction on the wheel is 2.55 N.m.

Step-by-step solution

Step 1: Given Data

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

Determining the concept

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:

$$\begin{gathered} {\text{V}}_{\mathrm{f}}^2={\mathrm{V}}_{\mathrm{i}}^2+2\mathrm{ax} \\ \mathrm{\alpha }=\frac{\mathrm{a}}{\mathrm{r}} \\ \mathrm{\tau }=\mathrm{I}\times \mathrm{\alpha } \end{gathered}$$

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

(a) Determining the linear acceleration

Magnitude of linear acceleration is given by the following formula:

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

Magnitude of acceleration is,

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

Hence, linear acceleration is 4.11 m/s².

(b) Determining the angular acceleration

Angular acceleration is given by the following formula:

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

Hence,angular acceleration is 16.4 rad/s².

(c) Determining the magnitude of the torque about the central axis due to friction on the wheel

Now, to findthetorque use the following formula:

$$\begin{gathered} \mathrm{\tau }=\mathrm{I}\times \mathrm{\alpha } \\ \mathrm{\tau }=0.155\times 16.4 \\ \mathrm{\tau }=2.542\mathrm{N}.\mathrm{m} \\ \mathrm{\tau }=2.55\mathrm{N}.\mathrm{m} \end{gathered}$$

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.