December 2, 2022 – Physicspupil

A frictionless circular track of radius $15.3\,\mathrm{cm}$ is fixed vertically on the floor. A particle at the top of the track glides down from rest.

Find the height at which the particle will begin to leave the circular track. Calculate the horizontal and vertical components of the velocity with which the particle strikes the floor.

Roughwork.

Set up a coordinate system:

with an interface:

Write resolved $x$ -, $y$ -components of normal reaction $N$

$\begin{aligned} N_x & = N\cos\theta = mg\sin 2\theta \\ N_y & = N\sin\theta =mg\sin^2\theta \\ \end{aligned}$

For the particle to lose contact with the surface, necessarily there exists some largest possible angle $\theta\in [0,90^\circ )$ s.t.

$N_x(\theta )\textrm{ \scriptsize{OR} }N_y(\theta )=0$ ;

and sufficiently some smallest possible period $t\in \Big[0, \sqrt{\frac{2R}{g}}\Big)$ s.t.

$s_x^2(t)+s_y^2(t)>R^2$

hereby SUVAT equations of motion do $\textrm{\scriptsize{NOT}}$ apply because of non-uniform acceleration $a(\theta )$ depending on $\theta$ , e.g.,

$\begin{aligned} \textrm{Net }F_x = ma_x & = N_x \\ a_x(\theta ) & = g\sin 2\theta \\ \textrm{Net }F_y = ma_y & = W-N_y \\ a_y & = g-g\sin^2\theta \\ a_y(\theta ) & = g\cos^2\theta \\ \end{aligned}$

By quotient rule,

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\theta}\bigg(\frac{s_y(\theta)}{s_x(\theta )}\bigg) & = \frac{\mathrm{d}}{\mathrm{d}\theta} (\tan\theta )\\ \frac{s_x(\theta)v_y(\theta)-s_y(\theta)v_x(\theta)}{s_x^2(\theta)} & = \sec^2\theta \\ \end{aligned}$

Try considering the Lagrangian $\mathcal{L}=T-V$ by

$\begin{aligned} T & = \frac{1}{2}m(\dot{s_x}^2+\dot{s_y}^2) \\ V & = mgs_y \\ \end{aligned}$

will not work. Try instead mathematically, first by noting $s=\sqrt{s_x^2+s_y^2}$ and $\theta = s/R$ . On one hand, along $s$ we have one equation of motion with initial boundary conditions:

$\begin{aligned} 0 & =\frac{\mathrm{d}^2s}{\mathrm{d}t^2}-g\cos\bigg(\frac{s}{R}\bigg) \\ 0 & = \bigg[\frac{\mathrm{d}s}{\mathrm{d}t}\bigg]\bigg|_{t=0} \\ \frac{\pi R}{2} & = s(0) \\ \end{aligned}$

on the other hand, along line of action of the centripetal force,

$\begin{aligned} & F_\textrm{C} = \frac{m\dot{s}^2}{R} = mg\sin \bigg( \frac{s}{R} \bigg) \\ & \bigg(\frac{\mathrm{d}s}{\mathrm{d}t}\bigg)^2 - gR\sin\bigg(\frac{s}{R}\bigg) = 0 \\ \end{aligned}$

we have one another. Then,

$\begin{aligned} \frac{\mathrm{d}s}{\mathrm{d}t}\frac{\mathrm{d}^2s}{\mathrm{d}t^2}-g\frac{\mathrm{d}s}{\mathrm{d}t}\cos\bigg(\frac{s}{R}\bigg) & = 0 \\ \frac{\mathrm{d}s}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\mathrm{d}s}{\mathrm{d}t}\bigg) -\frac{\mathrm{d}}{\mathrm{d}t}\bigg[ gR\sin\bigg(\frac{s}{R}\bigg)\bigg] & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg[ \frac{1}{2}\bigg(\frac{\mathrm{d}s}{\mathrm{d}t}\bigg)^2 - gR\sin \bigg(\frac{s}{R}\bigg) \bigg] & = 0 \\ \end{aligned}$

This problem is not to be attempted.

(discontinued)

A rectangular loop of length $0.2\,\mathrm{m}$ and width $0.1\,\mathrm{m}$ , carrying a steady current $I$ of $2\,\mathrm{A}$ is hinged along the $y$ -axis, and is situated in a uniform magnetic field $\mathbf{B}$ of $0.5\,\mathrm{T}$ parallel to the $x$ -axis.

If the plane of the loop makes an angle of $60^\circ$ with the $x$ – $y$ plane,

(a) calculate the force exerted by the magnetic field on each side of the loop, and
(b) calculate the torque required to hold the loop in this position.

Roughwork.

Force on a current-carrying conductor in a magnetic field is in magnitude

$\boxed{F=BIl\sin\theta}$