Discussion Forum

Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=at^{2}$ , then the  total acceleration of a point A on the rim of the wheel is (v being the tangential velocity)

Answer:

Explanation:

Given , $\theta=at^{2}$

 so  angular velocity ,$\omega=\frac{d\theta}{dt}=2at$

 linear tangential velocity,

 $v=\omega r=2atr $       $\frac{v}{t}=2ar$

So, tangential linear acceleration of particle is 

 $a_{1}=\frac{dv}{dt}=2ar$

and angular acceleration of partcle  is 

$\alpha=\frac{d\omega}{dt}=2a$

 Also, normal or radial acceleration of particle  is

$a_{n}=\frac{v^{2}}{r}= \frac{4a^{2}t^{2}r^{2}}{r}=4a^{2}t^{2}r$

Total acceleration of particle is

 $a_{total}=\sqrt{a_{t}^{2}+a_{n}^{2}}=\sqrt{4a^{2}r^{2}+16a^{4}t^{4}r^{2}}$

$=2ar\sqrt{1+4a^{2}t^{4}}$

$=\frac{v}{t}\sqrt{1+4a^{2}t^{4}}$