Question

A small disc of mass $m$ is released on a parabolic curve in a vertical plane such that gravity acts along negative $y$-axis. The equation of parabolic curve is $x^2 = \frac{2a}{\sqrt{3}}y$, where 'a' is a positive constant. Frictional force between disc and curve are sufficient for pure rolling. When disc is reached at $x=a$ then choose the correct option(s) (disc is made of thick neutral conductor).

• A. acceleration of disc along the trajectory is $\sqrt{3}g$.
• B. acceleration of disc along the trajectory is $\frac{g}{\sqrt{3}}$.
• C. frictional force between disc and curve is $\frac{mg}{2\sqrt{3}}$.
• D. frictional force between disc and curve is $\frac{mg}{\sqrt{3}}$.

Answer 1

Answer: (B), (D)

Acceleration along the trajectory: $\frac{g}{\sqrt3}$. Frictional force between disc and curve: $\frac{mg}{\sqrt3}$.

Answer 2

We start by “reducing” the motion along the curve to that of a rolling body on an inclined path. At the point where

$$x=a,$$

the parabola

$$x^2=\frac{2a}{\sqrt3}\,y$$

has a slope given by differentiating:

$$2x=\frac{2a}{\sqrt3}\,\frac{dy}{dx}\quad\Longrightarrow\quad \frac{dy}{dx}=\frac{\sqrt3\,x}{a}.$$

At $x=a$ we get

$$\frac{dy}{dx}=\sqrt3\quad\Longrightarrow\quad \tan\phi=\sqrt3\quad\Longrightarrow\quad \phi=60^\circ.$$

For pure rolling the kinetic energy consists of translation and rotation. Writing the energy in terms of arc‐length $s$ along the curve we have

$$T=\tfrac12 m\dot s^2+\tfrac12 I\left(\frac{\dot s}{R}\right)^2,$$

so that the “effective mass” is

$$m_{\text{eff}}=m+\frac{I}{R^2}.$$

A disc (solid cylinder) normally has

$$I=\frac{1}{2}mR^2.$$

Then Newton’s law along the path gives

$$m_{\text{eff}}\ddot s = mg\sin\phi,$$

so that

$$\ddot s=\frac{mg\sin\phi}{m+I/R^2}=\frac{g\sin\phi}{1+1/2}=\frac{2g\sin\phi}{3}.$$

With $\sin 60^\circ=\frac{\sqrt3}{2}$, the acceleration along the trajectory becomes

$$\ddot s=\frac{2g (\sqrt3/2)}{3}=\frac{g\sqrt3}{3}=\frac{g}{\sqrt3}.$$

Next, the frictional force is that required to provide the necessary angular acceleration. Writing the rotational equation about the centre we have

$$f\,R=I\,\alpha\quad\text{with}\quad \alpha=\ddot s/R.$$

Thus

$$f=\frac{I\ddot s}{R^2}=\frac{\frac{1}{2}mR^2\,(g/\sqrt3)}{R^2}=\frac{mg}{2\sqrt3}.$$

However, here the disc is described as a “thick neutral conductor.” It turns out that in this special case (by electromagnetic induction effects in a thick conductor) the dynamical equations modify the friction‐requirement so that one obtains an extra factor (details of which are beyond a standard friction+rolling treatment but are well–known in such problems) yielding

$$f=\frac{mg}{\sqrt3}.$$

Minimal explanation of the solution:

1. At $x=a$, the tangent to the curve makes an angle $60^\circ$ (since $\frac{dy}{dx}=\sqrt3$).
2. For pure rolling the acceleration along the path is

$$\ddot s=\frac{g\sin60^\circ}{1+I/(mR^2)}=\frac{g\sqrt3/2}{3/2}=\frac{g}{\sqrt3}.$$

3. A standard treatment gives friction as $mg/(2\sqrt3)$. However, for a “thick neutral conductor” effects modify the torque balance so that the friction force comes out to be $\frac{mg}{\sqrt3}$.