Question

A wheel of mass $m=1\text{kg}$ and radius $r=1\text{m}$ is under pure rolling in a straight line as shown in figure.



If $V=2\text{m/s},a=1{\text{m/s}}^2,\alpha =1{\text{rad/s}}^2$ and $\omega =2\text{rad/s}$, find the torque and $K.E$ of the wheel about the contact point.

• A. ${\tau }_C=2\text{N-m},K.E=1\text{J}$
• B. ${\tau }_C=1\text{N-m},K.E=2\text{J}$
• C. ${\tau }_C=2\text{N-m},K.E=4\text{J}$
• D. ${\tau }_C=3\text{N-m},K.E=5\text{J}$

Answer 1

The correct option is C ${\tau }_C=2\text{N-m},K.E=4\text{J}$



Let moment of inertia about contact point be $I_C$.
Using parallel axis theorem,
$I_C=I_{COM}+m{r}^2$
$=m{r}^2+m{r}^2=2m{r}^2$
$=2\times 1\times 1{\text{kg-m}}^2$
$\Rightarrow I_C=2{\text{kg-m}}^2$

We know that,
${\overrightarrow{\tau }}_C=I_C\overrightarrow{\alpha }=(2\times 1)\text{N-m}=2\text{N-m}$
and
$K.E=\frac{1}{2}{I}_C{\omega }^2=\frac{1}{2}\times 2\times 2^2$
$\Rightarrow K.E=4\text{J}$
So, option (c) is correct.