Uniform Circular Motion — Question 3

Question 3

A curvature limit constrains speed quadratically

A test cart follows a circular path of radius 50 m at constant speed. Its permitted acceleration magnitude is at most 8 m/s$^2$. Use only the kinematic relation $a=v^2/R$.

1. Find the largest allowed speed and the corresponding period.

2. If speed is doubled at the same radius, find the acceleration and compare it with the limit.

3. What radius would permit twice the original maximum speed while retaining the same acceleration limit?

Question 3 - Solution

A curvature limit constrains speed quadratically

(a) Solve the inequality before taking a square root.

$$v^2\le a_{\max}R=8(50)=400\ \Rightarrow\ \boxed{v_{\max}=20\,\mathrm{m/s}},\quad T=2\pi(50)/20=5\pi\,\mathrm s$$ $v^2\leq a_{\max}R=400$ m$^2$/s$^2$, so $v_{\max}=20$ m/s. At this speed the period is $2\pi(50)/20=5\pi\approx15.708$ s. Lower speeds have longer periods.

(b) Apply the square dependence.

$$a'=(2v)^2/R=4a=32\,\mathrm{m/s^2}>8\,\mathrm{m/s^2}$$ At 40 m/s and R=50 m, acceleration is $1600/50=32$ m/s$^2$, four times the limit. Doubling speed doubles the distance traveled per time but quadruples the rate of directional change measured as acceleration at fixed radius.

(c) Increase the radius by the required factor.

$$R'\ge\frac{40^2}{8}=200\,\mathrm m=4R$$ The minimum new radius is $R=v^2/a_{\max}=1600/8=200$ m. It is four times the original radius, not twice. This calculation imposes a kinematic constraint without specifying the physical mechanism that supplies the acceleration.