Answer
1. Figure 15 (A)
The particle is slowing down.
2. Figure 15 (B)
The particle is speeding up.
3. Figure 15 (C)
The speed is momentarily zero.
Work Step by Step
Let ${\bf{a}}$ denote the acceleration vector of the particle moving along the circle as is shown in Figure 15. Let $v\left( t \right)$ be the speed of the particle. By Eq. (1), we have
${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$,
${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = v'\left( t \right)$,
${a_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$
Notice that ${a_{\bf{N}}} \ge 0$ but ${a_{\bf{T}}}$ is positive or negative, depending on whether the object is speeding up or slowing down.
1. Figure 15 (A)
From the figure we notice that the angle between ${\bf{a}}$ and ${\bf{T}}$ is obtuse, so ${\bf{a}}\cdot{\bf{T}} = {a_{\bf{T}}} < 0$. Hence, the particle is slowing down.
2. Figure 15 (B)
The angle between ${\bf{a}}$ and ${\bf{T}}$ is acute, so ${\bf{a}}\cdot{\bf{T}} = {a_{\bf{T}}} > 0$. Hence, the particle is speeding up.
3. Figure 15 (C)
From the figure we notice that ${\bf{a}}$ and ${\bf{T}}$ are parallel. So, the normal component of the acceleration vanishes, that is, ${a_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2} = 0$. Since $\kappa \left( t \right) = \frac{1}{R} \ne 0$, the speed is momentarily zero.
