Question

A wheel rotates around a stationary axis so that the rotation angle $\phi$ varies with time as $\phi =a{t}^2$, where $a=0.20rad/s^2$. the total acceleration $\omega$ of the point $A$ at the rim at the moment $t=2.5s$ if the linear velocity of the point $A$ at this moment $v=0.65m/s$. Find $10\omega$.

Answer 1

Here, angle is, $\phi =a{t}^2$

So, angular speed is, $\Omega =\frac{d\phi }{dt}=2at=2\times 0.2t=0.4t$

and, angular accelration is, $\alpha =\frac{d\Omega }{dt}=2a=0.4$ $rad/s^2$

So linear velocity is, $v=\Omega R⟹0.4tR=0.65$

At $t=2.5s$,

$0.4\times 2.5\times R=0.65⟹R=0.65$ $m$

So tangential acceleration is $a_t=\alpha R=0.4\times 0.65=0.26m/s^2$

Radial acceleration is $a_r=v^2/R=(0.65{)}^2/0.65=0.65m/s^2$

Net acceleration is $\omega =\sqrt{{0.26}^2+{0.65}^2}=0.7$

So, $10\omega =7$