Question

The COM of a uniform disc has velocity 4i on a horizontal plane The angular velocity of this disc is $\sqrt{5}$ (-k). The radius of curvature of the leftmost point of the disc at this instant is. Radius of disc is 8/$\sqrt{5}$

Answer 1

5

Answer 2

Let the velocity of the center of mass (COM) of the disc be $\vec{v}_C = 4\hat{i}$.

The angular velocity of the disc is $\vec{\omega} = \sqrt{5}(-\hat{k}) = -\sqrt{5}\hat{k}$.

The radius of the disc is $R = \frac{8}{\sqrt{5}}$.

Let the origin be at the COM. The leftmost point P of the disc is located at $\vec{r}_{CP} = -R\hat{i} = -\frac{8}{\sqrt{5}}\hat{i}$ relative to the COM.

The velocity of point P is given by $\vec{v}_P = \vec{v}_C + \vec{\omega} \times \vec{r}_{CP}$.

$\vec{v}_P = 4\hat{i} + (-\sqrt{5}\hat{k}) \times (-\frac{8}{\sqrt{5}}\hat{i})$

$\vec{v}_P = 4\hat{i} + (\sqrt{5} \times \frac{8}{\sqrt{5}})(\hat{k} \times \hat{i})$

$\vec{v}_P = 4\hat{i} + 8\hat{j}$.

The speed of point P is $v_P = |\vec{v}_P| = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.

To find the radius of curvature $\rho$, we need the acceleration of point P. The radius of curvature is given by $\rho = \frac{v_P^2}{a_n}$, where $a_n$ is the component of acceleration perpendicular to the velocity.

The acceleration of point P is given by $\vec{a}_P = \vec{a}_C + \vec{\alpha} \times \vec{r}_{CP} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP})$.

Assuming the velocity of the COM and the angular velocity are constant at this instant, we have $\vec{a}_C = \vec{0}$ and $\vec{\alpha} = \vec{0}$.

So, $\vec{a}_P = \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP})$.

This term is the centripetal acceleration of P relative to C, given by $\vec{a}_{P/C}^{centripetal} = -\omega^2 \vec{r}_{CP}^{\perp}$, where $\vec{r}_{CP}^{\perp}$ is the component of $\vec{r}_{CP}$ perpendicular to $\vec{\omega}$. Since $\vec{\omega} = -\sqrt{5}\hat{k}$ is along the z-axis and $\vec{r}_{CP} = -\frac{8}{\sqrt{5}}\hat{i}$ is in the xy-plane, $\vec{r}_{CP}$ is perpendicular to $\vec{\omega}$.

$\vec{a}_P = -|\vec{\omega}|^2 \vec{r}_{CP} = -(\sqrt{5})^2 (-\frac{8}{\sqrt{5}}\hat{i}) = -5 (-\frac{8}{\sqrt{5}}\hat{i}) = 5 \times \frac{8}{\sqrt{5}}\hat{i} = \frac{40}{\sqrt{5}}\hat{i} = 8\sqrt{5}\hat{i}$.

So, the acceleration of point P is $\vec{a}_P = 8\sqrt{5}\hat{i}$.

The radius of curvature is $\rho = \frac{v_P^2}{a_n}$. We need to find the normal component of acceleration $a_n$.

The normal acceleration $a_n$ is the component of $\vec{a}_P$ perpendicular to $\vec{v}_P$. The tangential acceleration $a_t$ is the component of $\vec{a}_P$ parallel to $\vec{v}_P$.

$a_t = \frac{\vec{a}_P \cdot \vec{v}_P}{|\vec{v}_P|}$.

$\vec{a}_P \cdot \vec{v}_P = (8\sqrt{5}\hat{i}) \cdot (4\hat{i} + 8\hat{j}) = (8\sqrt{5} \times 4) + (8\sqrt{5} \times 8) \times (\hat{i} \cdot \hat{j}) = 32\sqrt{5} + 0 = 32\sqrt{5}$.

$v_P = 4\sqrt{5}$.

$a_t = \frac{32\sqrt{5}}{4\sqrt{5}} = 8$.

The magnitude of the total acceleration is $a_P = |\vec{a}_P| = |8\sqrt{5}\hat{i}| = 8\sqrt{5}$.

The normal acceleration $a_n$ is related to the total acceleration $a_P$ and tangential acceleration $a_t$ by $a_P^2 = a_t^2 + a_n^2$.

$(8\sqrt{5})^2 = 8^2 + a_n^2$

$64 \times 5 = 64 + a_n^2$

$320 = 64 + a_n^2$

$a_n^2 = 320 - 64 = 256$

$a_n = \sqrt{256} = 16$.

Now we can calculate the radius of curvature:

$\rho = \frac{v_P^2}{a_n} = \frac{(4\sqrt{5})^2}{16} = \frac{80}{16} = 5$.

The radius of curvature of the leftmost point of the disc at this instant is 5.