Spring, Friction Patch, and a Vertical Loop
A small block of mass $m$ is pressed against a spring of force constant $k$, compressing it by
a distance $d$ on a smooth horizontal floor. When released, the block crosses a rough
strip of length $L$ (kinetic friction coefficient $\mu$) and then enters a smooth vertical
circular loop of radius $R$ that is tangent to the floor.
1. Find the minimum compression $d_{\min}$ for which the block just completes the loop.
2. For $d = d_{\min}$, find the normal reaction $N$ on the block at the lowest point of the loop.
3. If the block enters the loop too slowly to complete it, it leaves the track where $N = 0$.
Find the angle $\theta$ (measured at the centre from the upward vertical) of that point in
terms of the entry speed $v_{\text{bot}}$.
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Answer
$ d_{\min} = \sqrt{\dfrac{mg}{k}\,(2\mu L + 5R)} $
Solution
(a) Minimum compression. Let the loop (radius $R$) be smooth. To just complete the
loop, gravity alone supplies the centripetal force at the top:
$$mg = \frac{m v_{\text{top}}^2}{R} \;\Rightarrow\; v_{\text{top}}^2 = gR.$$
Energy from release to the top (height $2R$), losing $\mu m g L$ to friction on the rough
patch:
$$\tfrac{1}{2}k d_{\min}^2 = \mu m g L + \tfrac{1}{2}m v_{\text{top}}^2 + mg(2R)
= \mu m g L + \tfrac{5}{2}mgR.$$
$$\boxed{\,d_{\min} = \sqrt{\tfrac{mg}{k}\,(2\mu L + 5R)}\,}.$$
(b) Normal force at the lowest point. For $d = d_{\min}$, energy from release to the
bottom of the loop gives $\tfrac{1}{2}m v_{\text{bot}}^2 = \tfrac{1}{2}k d_{\min}^2 - \mu mgL
= \tfrac{5}{2}mgR$, so $v_{\text{bot}}^2 = 5gR$. At the bottom,
$$N - mg = \frac{m v_{\text{bot}}^2}{R} = 5mg \;\Rightarrow\; \boxed{N = 6mg}.$$
(c) Leaving the track. If the block enters the loop with a smaller speed $v_{\text{bot}}$
(where $2gR \le v_{\text{bot}}^2 < 5gR$) it leaves the track on the way up, where $N = 0$.
At an angle $\theta$ from the top, $mg\cos\theta = \dfrac{mv^2}{R}$ and energy conservation
on the smooth loop give
$$\boxed{\cos\theta = \frac{v_{\text{bot}}^2 - 2gR}{3gR}},\qquad
v_{\text{bot}}^2 = \frac{k d^2}{m} - 2\mu g L.$$