Relating the Linear and Angular Variables — Question 10

Question 10

Release removes the inward acceleration constraint

A small pellet is carried at the rim of a disk of radius $0.50\,\mathrm m$, rotating counterclockwise at $4\,\mathrm{rad/s}$ in a horizontal plane. It is released at $(0.50,0)$ with no impulse. After release neglect all horizontal forces.

1. Find its horizontal position $0.20\,\mathrm s$ later.

2. Compare with the position it would have occupied if still attached.

Question 10 - Solution

Release removes the inward acceleration constraint

(a) Derivation and calculation

No impulse means velocity is continuous across release: $$\vec v_0=4\hat z\times0.50\hat x=2\hat y\ \mathrm{m/s}.$$ With zero horizontal force after release, $$\vec r(t)=\vec r_0+\vec v_0t\Rightarrow \boxed{\vec r(0.20)=(0.50,0.40)\ \mathrm m}.$$

(b) Result and interpretation

If attached, the point would remain on the circle: $$\Delta\theta=\omega t=4(0.20)=0.80\ \mathrm{rad}.$$ $$\boxed{\vec r_{\mathrm{attached}}=(0.50\cos0.80,0.50\sin0.80) =(0.3484,0.3587)\ \mathrm m}.$$ The released path is the tangent line because the inward acceleration constraint is removed.

Physical check

The released point has distance $\sqrt{0.50^2+0.40^2}>0.50$ m from the axis.