Problem 278
Medium Translational Dynamics

A cart of mass m starts from rest at height h and runs along…

A cart of mass $m$ starts from rest at height $h$ and runs along a track ABC that turns at B into a vertical semicircle of radius $R$ (B at the bottom, C at the top). Friction is negligible. (a) What is the smallest $h$ for which the cart stays on the track up to C? (b) How does the normal force depend on the position on the semicircle?

Explanation of Method

Smallest starting height for a loop.

Step 1: (a) Condition at the top C

\[ N_C + mg = \frac{mv_C^2}{R}, \quad N_C \ge 0 \;\Longrightarrow\; v_C^2 \ge gR. \]

\[ mg\,(h - 2R) = \tfrac12 mv_C^2 \ge \tfrac12 mgR. \]

\[h_{\min} = \tfrac52 R\]

Step 2: (b) Normal force at angle $\varphi$ from the top

A point D at angle $\varphi$ from C is at height $R(1 + \cos\varphi)$: $v_D^2 = 2g\,\big(h - R(1 + \cos\varphi)\big)$.

\[ N_D = \frac{mv_D^2}{R} - mg\cos\varphi. \]

\[N(\varphi) = mg\left(\frac{2h}{R} - 2 - 3\cos\varphi\right)\]

For $h = \tfrac52 R$: $N(\varphi) = 3mg\,(1 - \cos\varphi)$, which is zero at the top and $6mg$ at the bottom B.

$C$ $B$ $D$ $\varphi$ $R$ $h = \tfrac52 R$ $A$ $N(\varphi) = mg\left(\tfrac{2h}{R} - 2 - 3\cos\varphi\right)$
The angle $\varphi$ is measured from the top C of the semicircle. The normal force is smallest at C; starting from $h = \tfrac52 R$ it drops to exactly zero there.

Result

\[\boxed{\,h_{\min} = \tfrac52 R\,}\]

\[\boxed{\,N(\varphi) = mg\left(\frac{2h}{R} - 2 - 3\cos\varphi\right)\,}\]

Remark

Check. The original results are correct. Note that $\varphi$ is measured from the top of the semicircle (the original also calls the top "$\varphi = 90^\circ$" in part (a), which is inconsistent with its own formula).