A cart of mass m is pushed at A with speed v_0 along a track…

A cart of mass $m$ is pushed at A with speed $v_0$ along a track ABC (A on the level of B), which at B turns into a vertical semicircle of radius $R$. What is the smallest initial kinetic energy for the cart to get round to C without leaving the track, if 2% of the energy is lost to friction? What is its speed at C?

Explanation of Method

Starting energy for a loop with friction losses.

Step 1: Energy needed at C

As in Translational Dynamics 117, the cart needs $v_C^2 \ge gR$ at the top: $E_C = 2mgR + \tfrac12 mv_C^2 = \tfrac52 mgR$.

Step 2: Including friction

If friction takes 2% of the initial energy, $0.98\,E_A = \tfrac52 mgR$: $E_{k,A} \approx 2.55\,mgR, \quad v_C = \sqrt{gR}$.

$C$ $B$ $D$ $\varphi$ $R$ $A$ $v_0$ $E_{k,A} \approx 1.02 \cdot \tfrac52 mgR = 2.55\,mgR$ $v_C = \sqrt{gR}$

The cart starts at A on the level of B with speed $v_0$. It needs $2mgR$ to climb to C, $\tfrac12 mgR$ to still have the speed $\sqrt{gR}$ there, and 2% extra for friction.

Result

\[\boxed{\,E_{k,A} \approx 2.55\,mgR, \quad v_C = \sqrt{gR}\,}\]

Remark

Check. The original result is correct. It added 2% to the needed energy ($1.02 \times 2.5 = 2.550$); taking 2% of the initial energy gives $2.5/0.98 = 2.551$, the same to three digits.