Question

A wheel rotates around a stationery axis so that rotation angle $\theta$ varies as $\theta ={pt}^2$, where $p=0.20{rad/s}^2$. Find the total acceleration a of the point A at the rim at the moment t=2.55 sec, if the linear velocity of the point A at this moment is v=0.65 m/s.

Answer 1

$\theta =p{t}^2,v=0.65m/s,t=2.55sec,p=0.20rad/s$

Differentiating $\theta \left(t\right)=w_z=2pt$ _________ (1)

for fixed rotation ; speed of point A

$v=w_{z}R=2ptR$ or $R=\frac{v}{2pt}$ ________________ (2)

Differentiating w.r.t time

$w_t=\frac{dv}{dt}=2pR=\frac{v}{t}$ (using 1)

But = $w_r=\frac{v^2}{R}=\frac{v^2}{v/2pt}=2ptv$ (using 2)

$w=\sqrt{w_t^2}=\sqrt{(v/t{)}^2+(2atv{)}^2}=\frac{v}{t}\sqrt{1+4a^2{t}^4}$

$=\frac{0.65}{2.55}\sqrt{1+(0.2{)}^2\times 4\times (2.55{)}^4}=0.71rad/s^2$