Relating the Linear and Angular Variables — Question 7

Question 7

A sliding bead is not a fixed-radius point of the rotor

A bead is constrained to a straight radial slot rotating at constant $\omega=3\,\mathrm{rad/s}$. Its radius is $r(t)=0.20+0.10t\,\mathrm m$. Use the polar acceleration relations $a_r=\ddot r-r\omega^2$ and $a_\theta=r\alpha+2\dot r\omega$.

1. Find velocity components and acceleration components at $t=1\,\mathrm s$.

2. Explain the error in using only $a_\theta=r\alpha$.

Question 7 - Solution

A sliding bead is not a fixed-radius point of the rotor

(a) Derivation and calculation

At $t=1$ s, the given radial law yields $$r=0.30\ \mathrm m,\quad\dot r=0.10\ \mathrm{m/s},\quad\ddot r=0,\quad\omega=3,\quad\alpha=0.$$ $$\boxed{(v_r,v_\theta)=(\dot r,r\omega)=(0.10,0.90)\ \mathrm{m/s}}.$$ $$a_r=0-0.30(3)^2=-2.70,\quad a_\theta=0+2(0.10)(3)=0.60, \quad\boxed{(a_r,a_\theta)=(-2.70,0.60)\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

The full polar formula separates two tangential contributions: $$a_\theta=\underbrace{r\alpha}_{0}+\underbrace{2\dot r\omega}_{0.60\ \mathrm{m/s^2}}.$$ The bead slides relative to the rotor, so $\dot r\ne0$ and the fixed-radius formula $a_\theta=r\alpha$ omits a required term.

Physical check

Setting the radial sliding rate to zero recovers the usual rigid-body formula.