Uniform Circular Motion — Question 1

Question 1

Centripetal force names the resultant, not another interaction

A 2 kg puck moves at constant speed 6 m/s on a frictionless horizontal table in a circle of radius 3 m. A light cord attached to the center provides the only horizontal force.

1. Find the cord tension and the net horizontal force vector at the rightmost point.

2. Find the period and explain why adding another inward force $mv^2/r$ would double-count.

3. If the cord breaks at that point while the puck moves upward in the plan view, find its position relative to the center 0.50 s later.

Question 1 - Solution

Centripetal force names the resultant, not another interaction

(a) Use the radial acceleration.

Tension supplies the entire inward resultant: $$\boxed{T=\frac{mv^2}{r}=\frac{2(6)^2}{3}=24\ \mathrm N}.$$ At the rightmost point, inward is negative $x$, so $\boxed{\vec F_{\mathrm{net}}=-24\hat{\mathbf i}\ \mathrm N}$.

(b) Relate one circumference to the speed.

One revolution covers $2\pi r$: $$\boxed{\mathcal T=\frac{2\pi r}{v}=\frac{2\pi(3)}6=\pi\ \mathrm s}.$$ The radial equation is $\sum F_r=T=mv^2/r$. Adding an extra force called $mv^2/r$ would count the same inward resultant twice.

(c) Remove the vanished force.

After the cord breaks, $\vec a=0$ in the table plane. Hence $$\vec r(t)=\vec r_0+\vec v_0t=(3,0)+(0,6)t.$$ $$\boxed{\vec r(.50)=(3,3)\ \mathrm m}.$$ The subsequent straight path is tangent to the circle at release.