Question

At time t = 0, a disk of diameter 4m starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha = \frac{2}{3}$ rad $s^{-2}$. A small stone is stuck to the disk. At t = 0, it is at the contact point of the disk and the plane. Later, at time t = $\sqrt{\pi}$ s, the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$. The value of x is ______. [Take g = 10 m $s^{-2}$]

Answer 1

7

Answer 2

The disk has a diameter of 4m, so its radius is $R = 2$m. The disk starts from rest ($\omega_0 = 0$) with an angular acceleration $\alpha = \frac{2}{3}$ rad $s^{-2}$. The stone is initially at the contact point of the disk and the plane at $t=0$. The disk rolls without slipping, so the velocity of the center of the disk is $v_{cm} = \omega R$ and the acceleration of the center of the disk is $a_{cm} = \alpha R$.

At time $t_d = \sqrt{\pi}$ s, the stone detaches. The angular velocity of the disk at time $t_d$ is $\omega_d = \omega_0 + \alpha t_d = 0 + \frac{2}{3} \sqrt{\pi} = \frac{2\sqrt{\pi}}{3}$ rad $s^{-1}$. The angular displacement of the disk at time $t_d$ is $\theta_d = \theta_0 + \omega_0 t_d + \frac{1}{2} \alpha t_d^2$. Assuming $\theta_0 = 0$ (initial angular position relative to the bottom), $\theta_d = 0 + 0 + \frac{1}{2} \left(\frac{2}{3}\right) (\sqrt{\pi})^2 = \frac{1}{3} \pi$ rad.

Let's set the origin of the coordinate system at the initial contact point (0, 0). The horizontal direction is the x-axis and the vertical direction is the y-axis. The center of the disk is initially at (0, R). The horizontal distance rolled by the disk at time $t_d$ is $s = R \theta_d = 2 \left(\frac{\pi}{3}\right) = \frac{2\pi}{3}$. The position of the center of the disk at time $t_d$ is $C = (s, R) = (\frac{2\pi}{3}, 2)$.

The stone is initially at the bottom of the disk (relative to the center, its position is $(0, -R)$). After rotating by $\theta_d = \frac{\pi}{3}$ counter-clockwise relative to the center, its position relative to the center is $(R \sin \theta_d, -R \cos \theta_d)$. Position of stone relative to center $\vec{r}_{s/c} = (2 \sin(\frac{\pi}{3}), -2 \cos(\frac{\pi}{3})) = (2 \frac{\sqrt{3}}{2}, -2 \frac{1}{2}) = (\sqrt{3}, -1)$.

The position of the stone at detachment is $\vec{r}_d = \vec{r}_c + \vec{r}_{s/c} = (\frac{2\pi}{3}, 2) + (\sqrt{3}, -1) = (\frac{2\pi}{3} + \sqrt{3}, 1)$. The height of the stone at detachment is $y_d = 1$ m.

The velocity of the stone at detachment is the vector sum of the velocity of the center of the disk and the velocity of the stone relative to the center. The velocity of the center of the disk at time $t_d$ is $\vec{v}_{cm} = (v_{cm}, 0) = (\omega_d R, 0) = \left(\frac{2\sqrt{\pi}}{3} \times 2, 0\right) = \left(\frac{4\sqrt{\pi}}{3}, 0\right)$. The velocity of the stone relative to the center is $\vec{v}_{rel} = \vec{\omega}_d \times \vec{r}_{s/c}$. The angular velocity vector is $\vec{\omega}_d = (0, 0, \omega_d)$ (counter-clockwise rotation). $\vec{v}_{rel} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & \omega_d \\ \sqrt{3} & -1 & 0 \end{array}\right| = \hat{i}(0 - (-\omega_d)) - \hat{j}(0 - \omega_d \sqrt{3}) + \hat{k}(0 - 0) = \omega_d \hat{i} + \sqrt{3} \omega_d \hat{j}$. $\vec{v}_{rel} = \left(\frac{2\sqrt{\pi}}{3}, \sqrt{3} \frac{2\sqrt{\pi}}{3}\right) = \left(\frac{2\sqrt{\pi}}{3}, \frac{2\sqrt{3}\sqrt{\pi}}{3}\right)$.

The velocity of the stone at detachment is $\vec{v}_d = \vec{v}_{cm} + \vec{v}_{rel} = \left(\frac{4\sqrt{\pi}}{3}, 0\right) + \left(\frac{2\sqrt{\pi}}{3}, \frac{2\sqrt{3}\sqrt{\pi}}{3}\right) = \left(\frac{6\sqrt{\pi}}{3}, \frac{2\sqrt{3}\sqrt{\pi}}{3}\right) = \left(2\sqrt{\pi}, \frac{2\sqrt{3}\sqrt{\pi}}{3}\right)$. The initial velocity components of the stone after detachment are $v_{dx} = 2\sqrt{\pi}$ and $v_{dy} = \frac{2\sqrt{3}\sqrt{\pi}}{3}$.

The stone flies off tangentially and follows a projectile motion under gravity. The initial height is $y_d = 1$ m. The maximum height reached by the stone is given by $H_{max} = y_d + \frac{v_{dy}^2}{2g}$. Given $g = 10$ m $s^{-2}$. $v_{dy}^2 = \left(\frac{2\sqrt{3}\sqrt{\pi}}{3}\right)^2 = \frac{4 \times 3 \times \pi}{9} = \frac{12\pi}{9} = \frac{4\pi}{3}$. $H_{max} = 1 + \frac{4\pi/3}{2 \times 10} = 1 + \frac{4\pi}{60} = 1 + \frac{\pi}{15}$.

The maximum height is given as $\frac{1}{2} + \frac{x}{10}$. So, $1 + \frac{\pi}{15} = \frac{1}{2} + \frac{x}{10}$. $1 - \frac{1}{2} + \frac{\pi}{15} = \frac{x}{10}$. $\frac{1}{2} + \frac{\pi}{15} = \frac{x}{10}$. Multiply by 30: $15 + 2\pi = 3x$. $x = \frac{15 + 2\pi}{3} = 5 + \frac{2\pi}{3}$.

Using $\pi \approx 3$: $x = 5 + \frac{2 \times 3}{3} = 5 + 2 = 7$.