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Collar sliding on rotating arm, acceleration

Short answer

θ=35° gives t=1.63 s. Then r=0.581 m, ṙ=−0.391 m/s, r̈=−0.24 m/s², θ̇=0.7497 rad/s, θ̈=0.46. a_r = r̈−rθ̇² = −0.567 m/s²; a_θ = rθ̈+2ṙθ̇ = −0.319 m/s²; total a = 0.650 m/s².

Rotation of the arm about O is defined by θ=0.23t2\theta = 0.23t^2 where θ\theta is in radians and tt in seconds. Collar B slides along the arm such that r=0.9−0.12t2r = 0.9 - 0.12t^2 where rr is in meters. After the arm has rotated through 35∘35^\circ, determine the total acceleration of the collar. [Figure: arm OA rotating about O at angle θ\theta, collar B at distance rr from O.]

Use polar coordinates with rr along the arm and θ\theta the arm rotation.

Time when θ=35∘\theta=35^\circ

θ=35∘=0.6109 rad=0.23t2 ⇒ t2=2.656, t=1.6297 s\theta=35^\circ=0.6109\ \text{rad}=0.23t^{2}\ \Rightarrow\ t^{2}=2.656,\ t=1.6297\ \text{s}

Kinematic quantities at this instant

r=0.9−0.12t2=0.9−0.12(2.656)=0.5813 mr˙=−0.24t=−0.3911 m/s,r¨=−0.24 m/s2θ˙=0.46t=0.7497 rad/s,θ¨=0.46 rad/s2\begin{aligned} r&=0.9-0.12t^{2}=0.9-0.12(2.656)=0.5813\ \text{m}\\ \dot r&=-0.24t=-0.3911\ \text{m/s},\qquad \ddot r=-0.24\ \text{m/s}^2\\ \dot\theta&=0.46t=0.7497\ \text{rad/s},\qquad \ddot\theta=0.46\ \text{rad/s}^2 \end{aligned}

Acceleration components

ar=r¨−rθ˙2=−0.24−0.5813(0.7497)2=−0.5667 m/s2aθ=rθ¨+2r˙θ˙=0.5813(0.46)+2(−0.3911)(0.7497)=−0.3190 m/s2a=ar2+aθ2=0.650 m/s2\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-0.24-0.5813(0.7497)^{2}=-0.5667\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0.5813(0.46)+2(-0.3911)(0.7497)=-0.3190\ \text{m/s}^2\\ a&=\sqrt{a_r^{2}+a_\theta^{2}}=0.650\ \text{m/s}^2 \end{aligned}

Both components are negative: the acceleration points towards O and against the sense of increasing θ\theta, at tan⁡−1(0.319/0.567)=29.4∘\tan^{-1}(0.319/0.567)=29.4^\circ from the line towards O.

Answer: ar=−0.567a_r=-0.567 m/s2^2, aθ=−0.319a_\theta=-0.319 m/s2^2, total a≈0.650a\approx0.650 m/s2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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