Question

A wheel rotates around a stationary axis passing through its centre so that the angle of rotation $\varphi$ varies with time as $\varphi =k{t}^2$ where $k=\sqrt{2}{\text{rad/s}}^2$. Find the magnitude of total acceleration of the point at the rim at time $t=1\text{s}$, if the tangential velocity of the point at that moment is $2\text{m/s}$.

• A. $6{\text{m/s}}^2$
• B. $3{\text{m/s}}^2$
• C. $1{\text{m/s}}^2$
• D. $2{\text{m/s}}^2$

Answer 1

The correct option is A $6{\text{m/s}}^2$

The angular dispalacement of the wheel is given as, $\varphi =k{t}^2$
$\Rightarrow$ Angular velocity of the wheel is $\omega =\frac{d\varphi }{dt}=\frac{d}{dt}\left(k{t}^2\right)=2kt$
& Angular acceleration of the wheel is $\alpha =\frac{d\omega }{dt}=\frac{d^2\varphi }{d{t}^2}=\frac{d}{dt}\left(2kt\right)=2k$

Tangential acceleration $\left(a_t\right)$ of a point on the rim of the wheel is
$a_t=R\alpha =\left(2k\right)R....\left(i\right)$
Centripetal acceleration $\left(a_c\right)$ of that point is
$a_c={\omega }^{2}R=(2kt{)}^{2}R=4k^2{t}^{2}R....(ii)$

If $v$ is the tangential speed of a point on the rim then
$v=R\omega$
Or, $v=\left(2kt\right)R$
$\therefore R=\frac{v}{2kt}$
Using this value of $R$ in tangential and centripetal acceleration, we get
$a_t=2k\left(\frac{v}{2kt}\right)=\frac{v}{t}$
& $a_c=4k^2{t}^2\left(\frac{v}{2kt}\right)=2ktv$

The magnitude of total acceleration $\left(a_T\right)$ of this point is given by the vector diagram as shown below.


$a_T=\sqrt{a_t^2+a_N^2}$
$a_T=\sqrt{{\left(\frac{v}{t}\right)}^2+(2kvt{)}^2}$
$\Rightarrow a_T=\sqrt{{\left(\frac{2}{1}\right)}^2+(2\times \sqrt{2}\times 2\times 1{)}^2}$
$\therefore a_T=6{\text{m/s}}^2$