Average Acceleration and Instantaneous Acceleration — Question 3

Question 3

Nonzero acceleration can coexist with constant speed

A particle follows $\vec r=R[\cos(\omega t)\hat\imath+\sin(\omega t)\hat\jmath]$, with $R=3$ m and $\omega=2$ s$^{-1}$.

1. Differentiate to find v and a, then show that speed is constant.

2. Evaluate v and a at zero and state their relative directions.

3. Compute $\vec v\cdot\vec a$ at any time and explain what the zero result means for speed.

Question 3 - Solution

Nonzero acceleration can coexist with constant speed

(a) Differentiate the rotating radius.

$$\vec v=R\omega(-\sin\omega t,\cos\omega t),\quad\vec a=-\omega^2\vec r,\quad\boxed{v=R\omega=6\,\mathrm{m/s}}$$ $\vec v=R\omega[-\sin(\omega t),\cos(\omega t)]$ and $\vec a=-R\omega^2[\cos(\omega t),\sin(\omega t)]$. Speed is $R\omega=6$ m/s and acceleration magnitude is $R\omega^2=12$ m/s$^2$, both constant even though the vectors rotate.

(b) Read the tangent and inward directions.

$$\vec v(0)=(0,6)\,\mathrm{m/s},\quad\vec a(0)=(-12,0)\,\mathrm{m/s^2},\quad\vec v\cdot\vec a=0$$ At zero, $\vec r=(3,0)$ m, $\vec v=(0,6)$ m/s and $\vec a=(-12,0)$ m/s$^2$. Velocity points upward along the tangent; acceleration points left toward the center. They are perpendicular.

(c) Separate speed change from directional change.

$$\frac{d(v^2)}{dt}=2\vec v\cdot\vec a=0\ \Rightarrow\ v=\text{constant}$$ The dot product is proportional to $\sin(\omega t)\cos(\omega t)-\cos(\omega t)\sin(\omega t)=0$. Since $d(v^2)/dt=2\vec v\cdot\vec a$, the speed stays constant. Perpendicular acceleration changes velocity direction rather than speed.