Spare a farthing?

The curve $y=x^4 m^{-3}$ describes the edge of the bowl. So, we can consider a coordinate system in which the curve or the edge bowl passes through the origin.

Since, the coin maintains a constant height $0.01m (y=0.01)$ the value of $x = 0.1 m$ (obtained from the curve equation).

Finding the tangent of the curve i.e $\frac {dy}{dx}=400x^3$ . But $\frac {dy}{dx}=\tan \theta$ , where $\theta$ is the angle made by the tangent at point on the edge of the curve with the horizontal. $(\frac {dy}{dx})_{x=0.1}=(\tan \theta )_{x=0.1}= 400(0.1)^3=0.4.$ .

But we know that if a body is in a uniform circular motion on an inclined circular plane then angular velocity of the body is given by $\omega = \frac {\sqrt{gr \tan \theta}}{r}$ , where $r$ is the radius of curvature of the path in which the body(coin) is rotating. $r=.01m$ and $g=9.8m/s$

Here, tangent obtained at $x=0.1m$ becomes the tangent of the inclined plane at point $(0.01,0.1)$ .

Therefore the angular velocity of the coin is given by, $\omega =\frac {\sqrt{(0.4).(9.8)(0.1)}}{0.1} = 6.2609 \approx 6.26$ rad/s.

Therefore, angular velocity of the coin is $6.26$ rad/s

y=100x^4 = 0.01 --> x=0.1 m which is the radious of the circular path that coin slides on.

Let's find the slope (m) of the bowl at y=0.01 m (namely x=0.1 m) by taking derivative;

m=y'=400x^3=400*(0.1)^3 --> m=0.4=tan(theta)

where theta is the inclination angle.

Finally, sin(theta) component of the weight must be equal to the cos(theta) component of the centeral force.

m w^2 r cos(theta) = m g sin(theta) --> w = sqrt(g tan(theta)/r)

where m and w are mass and angular velocity respectively. We get;

w = sqrt(9.8*0.4/0.1) --> w = 6.26 rad/s

At the point of contact, the forces on the coin (gravitational, centripetal, and normal) must balance. Without loss of generality, choose z=0. The gravitational force is mg in the negative y-direction; the centripetal is $\frac {m v^2}{r}$ in the x-direction. The normal force N is perpendicular to the surface of revolution y = 100 $x^4$ . y = .01 m and y = 100 $x^4 m^{-3} \Rightarrow$ x = r = .1 m. The tangent is the derivative y' = 400 $x^3$ . So y' = $\tan \phi$ = .4 at x = .1 m.

Balancing x- and y-components separately gives : N $\sin \theta$ = $\frac {m v^2}{r}$ and N $\cos \theta$ = mg, where $\theta$ = angle between the vertical and the normal force. Dividing these equations gives $\tan \theta = \frac{mv^2}{mgr}$ . Solving for v gives v = $\sqrt {gr \tan \theta}$ . $\phi = \theta$ , since the normal is perpendicular to the tangent, and the horizontal component is perpendicular to the vertical.

The angular velocity is $\frac {2 \pi}{\frac{2 \pi r}{v}} = v/r = \frac {\sqrt {gr \tan \theta}}{r} = \frac {\sqrt {9.8*.1*.4}}{.1}$ = 6.26 rad/s.

The forces acting on the coin are its weight $mg$ (directed downwards or vertical), normal force $N$ (directed perpendicular to the surface of the bowl and inwards) and the inertia force $\frac{mv^2}{r}$ (directed away from the center of the bowl, perpendicular to its axis of rotation or horizontal), where $m$ is the mass of the coin, $v$ is its velocity and $r$ is its distance from the axis of rotation. Summing forces along the vertical direction, letting $N$ be directed at angle $\theta$ from the vertical, gives $Ncos\theta=mg$ or $N=\frac{mg}{cos\theta}$ . Summing forces along the horizontal direction gives $Nsin\theta=\frac{mv^2}{r}$ . Substituting $N$ and omitting common terms, we have $v^2=rg\frac{sin\theta}{cos\theta}=rg\frac{dy}{dx}$ . Since $y=100x^4$ , then $\frac{dy}{dx}=400x^3$ . When $y=0.01$ , $x=0.1$ . Evaluating the derivative at this given point gives $\frac{dy}{dx}=0.4$ . This is equal to the slope of the tangent line to the bowl at the given point, or $\frac{sin\theta}{cos\theta}$ . Now, we have $v^2=0.4rg$ . The velocity $v$ and angular velocity $\omega$ are related by the equation, $v=r\omega$ . So $\omega^2=\frac{0.4g}{r}$ but $r$ is just equal to $x=0.1$ . Thus, $\omega=6.26 rad/s$ .

At y=0.01 m, x=0.1 m

let w be its angular velocity.

slope at (0.1, 0.01) , tan q=400 \times x^3=0.4

Now, since it is not up or down,so along the surface resultant force is 0. which gives

g sin q =(x \times w^2 ) cos q

w^2=(g tan q) / x

w^2=(9.8 \times 0.4) / 0.1 =39.2

which gives , w=6.26

hence required answer, w=6.26

Let $\omega$ be the angular velocity of the coin, and $R$ be the radius of its revolution.

Since the curve is symmetric about $x = 0$ , WLOG take $x \ge 0$ . On the curve $y = (100 \ \text{m}^{-3})x^4$ , if $y = 0.01 \ \text{m}$ then $x = 0.1 \ \text{m}$ . Thus, the radius of revolution of the coin is $R = 0.1 \ \text{m}$ .

Let $\theta$ be the angle of inclination at $x = 0.1 \ \text{m}$ . Then, $\tan \theta = \left.\dfrac{dy}{dx}\right|_{x = 0.1 \ \text{m}} = \left.(400 \ \text{m}^{-3})x^3\right|_{x = 0.1 \ \text{m}} = 0.4$ .

Draw a FBD of the coin. The weight of the coin is $W = mg$ (down). If the normal force has magnitude $N$ , then vertical and horizontal components of the normal force are $N_y = N\cos \theta$ (up) and $N_x = N\sin \theta$ (left).

The coin has no vertical acceleration. Thus, $N_y - W = ma_y = 0$ , i.e. $N\cos \theta = mg$ .

The coin's horizontal (centripetal) acceleration is $R\omega^2$ (left). Thus, $N_x = ma_x$ , i.e. $N \sin \theta = mR\omega^2$ .

Dividing the 2nd equation by the 1st yields $\tan \theta = \dfrac{R \omega^2}{g}$ .

Therefore, $\omega = \sqrt{\dfrac{g \tan \theta}{R}} = 6.26 \ \text{rad/s}$ .

Simply, notice that there are only 2 forces acting on the ring as it goes around the edge: the normal contact force, and the gravitational force.

However notice that the normal contact force is perpendicular to the tangent at y = 0.01 m.

Let the angle between the normal contact force and the x axis be $\theta '$ . Notice, that this angle is complimentary to the angle that the tangent makes to the x axis. This angle is $\theta$ .

By considering the force diagram:

$N \cos \theta ' = F_c$

$N \sin \theta ' = mg$

Conversely,

$N \sin \theta = F_c = m \omega ^2 r$ _ (1)

$N \cos \theta = mg$ _ _(2)

Also note that:

$\tan \theta = \frac {dy}{dx} = 400 x^3$ __ (3)

Solving these equations one will get,

$mg \tan \theta = m \omega ^2 r$

$g 400 x^3 = \omega ^2 x$

$\omega = (400 g x^2 )^{0.5}$

Note that since $y = 0.01 = 100 x^4$ , $x = 0.1$ , Therefore,

$\omega = (400 g 0.01)^{0.5} = 6.26$

and thus the answer is 6.26 rad/s

The two forces acting on the coin are gravity and the normal force from the bowl. If we denote $\theta$ to be the angle the bowl makes with the horizontal at position $x$ , then $F\cos\theta = mg$ since $\Sigma F_y = 0$ and $F\sin\theta = m\omega^2 x$ since $\Sigma F_x = F_{centr} = m\omega^2 x$ . Substituting $F = \frac{mg}{\cos\theta}$ into the second equation and rearranging, we find that $\omega = \sqrt{\frac{g\tan\theta}{x}}$ . Now we have to find $x$ and $\tan\theta$ . Since $y = 0.01 = 100x^4$ , $x = \sqrt[4]{\frac{0.01}{100}} = 0.1m$ . We also know that $\tan\theta = \frac{dy}{dx}$ , and by differentiation we find that $\frac{dy}{dx} = 400x^3 = 400(0.1)^3 = 0.4$ . Plugging in, we find that $\omega = \sqrt{\frac{(9.8)(0.4)}{0.1}} \approx \boxed{6.26 rad/s}$ .