Source: Principles of Physics
Problem Sets:
Problem
A particle undergoes uniform circular motion. At time $t_1$, its velocity is $\vec{v}_1 = v_x \hat{i} + v_y \hat{j}$. At a later time $t_2$, its velocity is $\vec{v}_2 = -\vec{v}_1$. The time interval is $\Delta t = t_2 - t_1$.
- What is the magnitude of the particle's centripetal acceleration?
- What is the particle's average acceleration during the time interval $\Delta t$?
[Q1] $a_c = \frac{\pi \sqrt{v_x^2 + v_y^2}}{\Delta t}$ [Q2] $\vec{a}_{avg} = -\frac{2(v_x \hat{i} + v_y \hat{j})}{\Delta t}$
The speed of the particle, $v$, is constant due to uniform circular motion:
$$v = |\vec{v}_1| = \sqrt{v_x^2 + v_y^2}$$The condition $\vec{v}_2 = -\vec{v}_1$ indicates the particle has traveled to a diametrically opposite point, which is half a revolution. Therefore, the period of motion is $T = 2\Delta t$.
The radius $r$ of the circular path is related to the speed and period by $v = 2\pi r / T$.
$$r = \frac{vT}{2\pi} = \frac{v(2\Delta t)}{2\pi} = \frac{v\Delta t}{\pi}$$[Q1] The magnitude of the centripetal acceleration $a_c$ is given by $a_c = v^2/r$.
$$a_c = \frac{v^2}{v\Delta t/\pi} = \frac{\pi v}{\Delta t}$$Substituting the expression for $v$:
$$a_c = \frac{\pi \sqrt{v_x^2 + v_y^2}}{\Delta t}$$[Q2] The average acceleration $\vec{a}_{avg}$ is the change in velocity over the time interval.
$$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}$$Substituting $\vec{v}_2 = -\vec{v}_1$:
$$\vec{a}_{avg} = \frac{-\vec{v}_1 - \vec{v}_1}{\Delta t} = \frac{-2\vec{v}_1}{\Delta t} = -\frac{2(v_x \hat{i} + v_y \hat{j})}{\Delta t}$$