A smooth wire is shaped into a circle of centre O and radius 0.8m. The wire is fixed in a vertical plane. A small bead P of mass 0.03 kg is threaded on the wire and is projected along the wire from the highest point with a speed of $4.2$ ms$^{-1}$. When OP makes an angle $\theta$ with the upward vertical the speed of P is $v$ ms$^{-1}$. Show that $v^2=33.32-15.68cos\theta $
When forming the conservation of energy equation for when the line OP makes an angle $\theta$, in addition to the equation for kinetic energy I wrote that GPE = $mgr(1-cos \theta)$, but doing this got me $+15.68$ instead of $-15.68$. The mark scheme says use $-mgr(1-cos\theta)$, but I don't understand why - I thought $mgr(1-cos\theta)$ works regardless of the position of the particle in the circular motion? Why does it need to be negative?
You are correct that the energy is the sum of kinetic and potential energy, and that this total energy should stay constant. $$ E=\frac{mv(t)^2}2+mgr(1-\cosθ(t)). $$ What is asked is a formula where $v(t)$ is computed from $v(0)$ and $θ(t)$. Additionally, we know that $θ(0)=0$. When solving $$ \frac{mv(0)^2}2=E=\frac{mv(t)^2}2+mgr(1-\cosθ(t)). $$ for the kinetic energy at time $t$, the potential term is subtracted from both sides.