Disc Partitioned - 2

I would like to provide a solution to this fantastic question !!! PLEASE UPVOTE if you like it !!

oh i got it but a REAL LENGTHY ONE.

• you just need to find the ratio of moment of inertia of two parts about IAR ie instantaneous axis of rotation in this case the bottom most point.

• just found the moment of inertia of lower most part and for the inertia of uppermost subtract it from total.

• to find the the moment of inertia of the lower most part ,i found the moment of inertia of minor segments most challenging,i found moment of inertia about the center of circle,found the cm of the the segment and then finally found it about the bottom most point.

This isn't a solution but if you have attempted this question. Please share your line of thought and method!

Since no one has done a LaTeX solution yet, I thought I would write one up!

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First, we need to determine the section of the disc that has velocity less than $V$ .

We note that since the disc is rolling without slipping, each point on the disc would be moving with velocity $V$ , if the disc was not rotating. In addition, we know that with spinning, the angular velocity, $\omega$ is given by $\omega = \frac{V}{R}$ .

Let $\theta$ be the angle between the vertical and the line extending from the centre to some point $(r, \ \theta)$ of the circle. For such a point, the translational velocity from rotation is given by $\omega r$ , so we break it into components, which gives us a general velocity vector:

$\vec{v}(r, \ \theta) = \begin{pmatrix} V \\ 0 \end{pmatrix} + \begin{pmatrix} -\omega r \cos \theta \\ \omega r \sin \theta \end{pmatrix} = \begin{pmatrix} V - \omega r \cos \theta \\ \omega r \sin \theta \end{pmatrix} = V \begin{pmatrix} 1 - \frac{r}{R} \cos \theta \\ \frac{r}{R} \sin \theta \end{pmatrix}$

We can then conclude that the magnitude of velocity at the point $(r, \ \theta)$ is the norm of this vector:

$v(r, \ \theta) = \sqrt{v_x^2 + v_y^2} = V \sqrt{\Big(1 - \frac{r \cos \theta}{R} \Big)^2 + \Big( \frac{r \sin \theta}{R} \Big)^2} = V \sqrt{1 + \frac{r^2}{R^2} - \frac{2 r \cos \theta}{R}}$

We are interested in points $(r, \ \theta)$ such that $v(r, \ \theta) < V$ :

$V \sqrt{1 + \frac{r^2}{R^2} - \frac{2 r \cos \theta}{R}} < V \ \Rightarrow \ \sqrt{1 + \frac{r^2}{R^2} - \frac{2 r \cos \theta}{R}} < 1 \ \Rightarrow \ \frac{r^2}{R^2} - \frac{2 r \cos \theta}{R} < 0 \ \Rightarrow \ 0 \leq r < 2R \cos \theta$

Clearly, it must also be true that $0 \leq r \leq R$ . We know that $\cos(\pi/3) = \cos(5\pi/3) = 1/2$ , so it follows that for angles ranging from $0$ to $\pi/3$ and from $5\pi/3$ to $2\pi$ , all of the points on the disc will travel with velocity less than $V$ .

We also note that $\cos(\pi/2) = \cos(3\pi/2) = 0$ , so from $\theta = \pi/3$ to $\theta = \pi/2$ and from $\theta = 3\pi/2$ to $\theta = 5\pi/3$ , the range of possible values of $r$ is given by $0 < r < 2R \cos \theta$ . This allows us to integrate across each infinitesimal contribution to the kinetic energy in these bounds. We must calculate two integrals, with each set of bounds. Note that we use the symmetry of the problem to simplify the integrals:

$K_0 = 2 \displaystyle\int_{0}^{\pi/3} \displaystyle\int_0^R \frac{1}{2} v(r, \ \theta)^2 dm = \displaystyle\int_{0}^{\pi/3} \displaystyle\int_0^R v(r, \ \theta)^2 \frac{M}{A} r dr d\theta = \frac{M V^2}{\pi R^2} \displaystyle\int_{0}^{\pi/3} \displaystyle\int_0^R r + \frac{r^3}{R^2} - \frac{2 r^2 \cos \theta}{R} dr d\theta$

$= \frac{M V^2}{\pi} \displaystyle\int_{0}^{\pi/3} \frac{3}{4} - \frac{2 \cos \theta}{3} d\theta$

We set up the second integral in a similar fashion, but using the second set of bounds:

$K_1 = \frac{M V^2}{\pi R^2} \displaystyle\int_{\pi/3}^{\pi/2} \displaystyle\int_{0}^{2R \cos \theta} r + \frac{r^3}{R^2} - \frac{2 r^2 \cos \theta}{R} dr d\theta = \frac{M V^2}{\pi} \displaystyle\int_{\pi/3}^{\pi/2} 2 \cos^2 \theta - \frac{4 \cos^4 \theta}{3} d\theta$

where we used the symmetry of the problem to simplify the integrals. We wish to find the total kinetic energy, $K = K_0 + K_1$ , so we have:

$K = \frac{M V^2}{\pi} \Big[ \displaystyle\int_{0}^{\pi/3} \frac{3}{4} - \frac{2 \cos \theta}{3} d\theta + \displaystyle\int_{\pi/3}^{\pi/2} 2 \cos^2 \theta - \frac{4 \cos^4 \theta}{3} d\theta \Big] \approx \frac{0.2894}{\pi} MV^2$

where we evaluated the final integral numerically. Finally, we note that the total kinetic energy of the rotating disc is given by the rotational kinetic energy, plus the translational kinetic energy:

$K_{\text{tot}} = \frac{1}{2} MV^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} MV^2 + \frac{1}{4} MV^2 = \frac{3}{4} MV^2$

so the kinetic energy of the other pat of the disc, $K'$ is equal to $K_{\text{tot}} - K$ . This allows us to find the final ratio between the kinetic energies of the components of the disc:

$R = \frac{K_{\text{tot}} - K}{K} = \frac{K_{\text{tot}}}{K} - 1 = \frac{\frac{3}{4} MV^2}{\frac{0.2894}{\pi} MV^2} - 1 = \frac{3 \pi}{(4)(0.2894)} - 1 \approx \boxed{7.14}$