A wheel rotates around a stationary axis so that the rotation angle $ heta$ varies with time as $

Question

A wheel rotates around a stationary axis so that the rotation angle $	heta$ varies with time as $	heta = kt^3$, where $k = \frac{1}{9} 	ext{ rad/s}^3$. Find the magnitude of the total acceleration (in $	ext{m/s}^2$) of a point $A$ on the rim at the moment $t = 2 	ext{ s}$, given that the linear velocity of point $A$ at this moment is $v = 6 	ext{ m/s}$.

Professor PC Thomas Classes · Answer

Given $	heta = \frac{1}{9}t^3$. Angular velocity $\omega = \frac{d	heta}{dt} = \frac{1}{3}t^2$. At $t=2$, $\omega = 4/3 	ext{ rad/s}$. Angular acc $\alpha = \frac{d\omega}{dt} = \frac{2}{3}t$. At $t=2$, $\alpha = 4/3 	ext{ rad/s}^2$. Radius $R = v/\omega = \frac{6}{4/3} = 4.5 	ext{ m}$. Tangential acc $a_t = \alpha R = (4/3)(4.5) = 6 	ext{ m/s}^2$. Normal acc $a_n = v\omega = 6(4/3) = 8 	ext{ m/s}^2$. Total acc $a = \sqrt{a_t^2 + a_n^2} = \sqrt{36 + 64} = 10 	ext{ m/s}^2$.