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Car on circular curve, angular rate

Short answer

Angular rate θ̇ = v/r = 30/400 = 0.075 rad/s. With constant speed, acceleration is purely centripetal: a = v²/r = 30²/400 = 2.25 m/s².

The car travels along the circular curve of radius r=400r = 400 m with a constant speed of v=30v = 30 m/s. Determine the angular rate of rotation θ˙\dot\theta of the radial line rr and the magnitude of the car's acceleration.

The car moves on a circle with constant speed, so the speed is entirely transverse: v=rθ˙v=r\dot\theta.

Angular rate

θ˙=vr=30400=0.075 rad/s\dot\theta=\frac{v}{r}=\frac{30}{400}=0.075\ \text{rad/s}

Acceleration

With rr constant and vv constant: r˙=r¨=0\dot r=\ddot r=0 and θ¨=0\ddot\theta=0.

ar=r¨−rθ˙2=−400(0.075)2=−2.25 m/s2aθ=rθ¨+2r˙θ˙=0\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-400(0.075)^{2}=-2.25\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=0 \end{aligned}

So the acceleration is purely centripetal, a=v2r=302400=2.25 m/s2a=\dfrac{v^2}{r}=\dfrac{30^2}{400}=2.25\ \text{m/s}^2, directed towards the centre.

Answer: θ˙=0.075\dot\theta=0.075 rad/s; a=2.25a=2.25 m/s2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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