Uniform Circular Motion — Question 10

Question 10

Leaving the circle preserves the instantaneous tangent velocity

A point travels counterclockwise at speed 10 m/s on a circle of radius 5 m centered at the origin. At position angle $30^\circ$ it leaves the constraint. For the following 0.20 s, its acceleration is stipulated to be zero.

1. Find the position and velocity at release.

2. Find its position 0.20 s later and its distance from the old center.

3. Explain why continuing along the circle or moving radially outward would contradict the stated post-release acceleration.

Question 10 - Solution

Leaving the circle preserves the instantaneous tangent velocity

(a) Take the tangent in the counterclockwise sense.

$$\vec r_0=5(\cos30^\circ,\sin30^\circ)=(5\sqrt3/2,5/2)\,\mathrm m,\quad\boxed{\vec v_0=(-5,5\sqrt3)\,\mathrm{m/s}}$$ Release position is $(5\sqrt3/2,5/2)$ m. The unit tangent is $(-\sin30^\circ,\cos30^\circ)$, so velocity is $(-5,5\sqrt3)$ m/s. It is perpendicular to the radius, not parallel to it.

(b) Continue with the release velocity.

$$\vec r=\vec r_0+.20\vec v_0=(5\sqrt3/2-1,5/2+\sqrt3)\,\mathrm m,\quad|\vec r|=\sqrt{25+4}=\sqrt{29}\,\mathrm m$$ With zero acceleration, displacement in 0.20 s is $(-1,\sqrt3)$ m. New position is $(5\sqrt3/2-1,5/2+\sqrt3)$ m. Since the displacement is perpendicular to the original radius, its distance from C is $\sqrt{5^2+2^2}=\sqrt{29}\approx5.385$ m.

(c) Respect the velocity-change requirement.

$$\vec a_{\mathrm{after}}=\vec0\ \Rightarrow\ \vec v=\vec v_0;\qquad a_{\mathrm{circle}}=v^2/R=20\,\mathrm{m/s^2}\ne0$$ Following the circle would still require inward acceleration $v^2/R=20$ m/s$^2$. An immediate radial departure would require a discontinuous velocity turn. Zero post-release acceleration instead preserves the tangent velocity, producing a straight tangent path whose distance from the old center grows.