Solution to Question 2
Method: Instantaneous Centre of Rotation (I.C.R.)
Since the strings are inextensible and fixed to the ceiling, the velocity of any point on the string along its length must be zero. Consequently, the velocity of the contact points on the disc ($A$ and $B$) must be perpendicular to the strings.
The intersection of the lines perpendicular to the velocities of points on a rigid body defines the Instantaneous Centre of Rotation (I.C.R.).
Since $\vec{v}_A \perp \text{String}_1$ and $\vec{v}_B \perp \text{String}_2$, the lines containing the strings themselves are the perpendiculars to the velocity vectors. Thus, their intersection point $I$ is the I.C.R.
Figure: Geometry of the Instantaneous Centre of Rotation (I). Note that $OA \perp IA$ and $OB \perp IB$.
Geometric Analysis
Let $O$ be the center of the disc and $r$ be its radius. Let the strings leave the disc at tangent points $A$ and $B$, intersecting at point $I$.
- Tangency: The radii $OA$ and $OB$ are perpendicular to the tangents $IA$ and $IB$ respectively ($\angle OAI = \angle OBI = 90^\circ$).
- Symmetry: $\Delta OAI$ and $\Delta OBI$ are congruent right-angled triangles (RHS congruence: Hypotenuse $IO$ is common, legs $OA=OB=r$).
- Angle at ICR: The angle between the strings is given by the external geometry: $\angle AIB = \theta + \phi = 67^\circ + 53^\circ = 120^\circ$.
- Bisector: Due to symmetry, the line $IO$ bisects $\angle AIB$. Thus, the semi-vertical angle is:
$$ \angle AIO = \frac{120^\circ}{2} = 60^\circ $$
Calculation
In the right-angled triangle $\Delta OAI$:
$$ \sin(\angle AIO) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OA}{IO} $$
$$ IO = \frac{r}{\sin(60^\circ)} $$
The velocity of the center of mass $v_0$ is related to the angular velocity $\omega$ and the distance from the I.C.R. ($IO$) by:
$$ v_0 = \omega \cdot IO $$
Substituting $IO$:
$$ v_0 = \omega \cdot \frac{r}{\sin(60^\circ)} $$
Substitute Values:
$\omega = 5.0 \, \text{rad/s}$
$r = 10\sqrt{3} \, \text{cm}$
$\sin(60^\circ) = \frac{\sqrt{3}}{2}$
$$ v_0 = 5 \cdot \frac{10\sqrt{3}}{\frac{\sqrt{3}}{2}} $$
$$ v_0 = 5 \cdot 10\sqrt{3} \cdot \frac{2}{\sqrt{3}} $$
$$ v_0 = 5 \cdot 20 = 100 \, \text{cm/s} $$
Magnitude of velocity of the centre of the disc is $100 \text{ cm/s}$.
Method: Instantaneous Centre of Rotation (I.C.R.)
Since the strings are inextensible and fixed to the ceiling, the velocity of any point on the string along its length must be zero. Consequently, the velocity of the contact points on the disc ($A$ and $B$) must be perpendicular to the strings.
Since $\vec{v}_A \perp \text{String}_1$ and $\vec{v}_B \perp \text{String}_2$, the lines containing the strings themselves are the perpendiculars to the velocity vectors. Thus, their intersection point $I$ is the I.C.R.
Figure: Geometry of the Instantaneous Centre of Rotation (I). Note that $OA \perp IA$ and $OB \perp IB$.
Geometric Analysis
Let $O$ be the center of the disc and $r$ be its radius. Let the strings leave the disc at tangent points $A$ and $B$, intersecting at point $I$.
- Tangency: The radii $OA$ and $OB$ are perpendicular to the tangents $IA$ and $IB$ respectively ($\angle OAI = \angle OBI = 90^\circ$).
- Symmetry: $\Delta OAI$ and $\Delta OBI$ are congruent right-angled triangles (RHS congruence: Hypotenuse $IO$ is common, legs $OA=OB=r$).
- Angle at ICR: The angle between the strings is given by the external geometry: $\angle AIB = \theta + \phi = 67^\circ + 53^\circ = 120^\circ$.
- Bisector: Due to symmetry, the line $IO$ bisects $\angle AIB$. Thus, the semi-vertical angle is: $$ \angle AIO = \frac{120^\circ}{2} = 60^\circ $$
Calculation
In the right-angled triangle $\Delta OAI$: $$ \sin(\angle AIO) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OA}{IO} $$ $$ IO = \frac{r}{\sin(60^\circ)} $$
The velocity of the center of mass $v_0$ is related to the angular velocity $\omega$ and the distance from the I.C.R. ($IO$) by: $$ v_0 = \omega \cdot IO $$ Substituting $IO$: $$ v_0 = \omega \cdot \frac{r}{\sin(60^\circ)} $$
Substitute Values:
$\omega = 5.0 \, \text{rad/s}$
$r = 10\sqrt{3} \, \text{cm}$
$\sin(60^\circ) = \frac{\sqrt{3}}{2}$
$$ v_0 = 5 \cdot \frac{10\sqrt{3}}{\frac{\sqrt{3}}{2}} $$ $$ v_0 = 5 \cdot 10\sqrt{3} \cdot \frac{2}{\sqrt{3}} $$ $$ v_0 = 5 \cdot 20 = 100 \, \text{cm/s} $$