Collar on arm: velocity and relative acceleration

Time when $\theta=45^\circ$

$$\theta=45^\circ=0.7854\ \text{rad}=0.75t^{2}\ \Rightarrow\ t^{2}=1.0472,\ t=1.0233\ \text{s}$$

Quantities at this instant

$$\begin{aligned} r&=1-0.3t^{2}=0.6858\ \text{m}\\ \dot r&=-0.6t=-0.6140\ \text{m/s},\qquad \ddot r=-0.6\ \text{m/s}^2\\ \dot\theta&=1.5t=1.5350\ \text{rad/s},\qquad \ddot\theta=1.5\ \text{rad/s}^2 \end{aligned}$$

(a) Total velocity

$$\begin{aligned} v_r&=\dot r=-0.6140\ \text{m/s}\\ v_\theta&=r\dot\theta=0.6858(1.5350)=1.0528\ \text{m/s}\\ v&=\sqrt{v_r^{2}+v_\theta^{2}}=1.219\ \text{m/s} \end{aligned}$$

The velocity makes $120.3^\circ$ with $\hat e_r$, that is, $59.8^\circ$ with the direction along the arm towards O.

(b) Total acceleration

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-0.6-0.6858(1.5350)^{2}=-2.216\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.6858(1.5)+2(-0.6140)(1.5350)=-0.856\ \text{m/s}^2\\ a&=\sqrt{2.216^{2}+0.856^{2}}=2.376\ \text{m/s}^2 \end{aligned}$$

(c) Acceleration relative to the arm

The collar moves along the arm, so relative to the arm (a rotating frame) its motion is rectilinear along $r$:

$$a_{rel}=\ddot r=-0.6\ \text{m/s}^2$$

That is, $0.6$ m/s$^2$ directed towards O along the arm.

Answer: (a) $v=1.219$ m/s; (b) $a=2.376$ m/s$^2$; (c) $a_{rel}=0.6$ m/s$^2$ towards O.