Problem Solution
Kinematics of a Rolling Spool
Problem Statement
A spool of a bobbin rolls without slipping on a horizontal track. The radius of the spool is \( r \). At some instant, the speeds of two diametrically opposite points on the spool are \( v_1 \) and \( v_2 \). Find the speed of the centre of the bobbin.
1. Key Concept: Instantaneous Center of Rotation
Point A is the contact point with the ground. For rolling without slipping, A is the Instantaneous Center of Rotation (I.C.R.). Velocity vectors are always perpendicular to the line connecting the point to the I.C.R.
2. Identify Velocities:
Let the angular velocity be \(\omega\). The speed of any point is \( v = \omega \cdot (\text{distance from A}) \).
Therefore:
$$ v_1 = \omega \cdot AB \quad \text{and} \quad v_2 = \omega \cdot AD $$
$$ \Rightarrow AB = \frac{v_1}{\omega}, \quad AD = \frac{v_2}{\omega} $$
3. Geometry (Right Angled Triangle):
Since \( B, C, D \) form a diameter of the circle, and \( A \) lies on the circle, the triangle \( \Delta BAD \) is a right-angled triangle (Angle in a semicircle is \( 90^\circ \)).
4. Apply Pythagoras Theorem:
In \( \Delta BAD \):
$$ AB^2 + AD^2 = BD^2 $$
Since \( BD \) is the diameter, \( BD = 2r \).
$$ \left(\frac{v_1}{\omega}\right)^2 + \left(\frac{v_2}{\omega}\right)^2 = (2r)^2 $$
5. Solve for Center Velocity:
Multiplying by \( \omega^2 \):
$$ v_1^2 + v_2^2 = \omega^2 (4r^2) = 4 (\omega r)^2 $$
The velocity of the center \( C \) is given by \( v_c = \omega r \). Substituting this back:
$$ v_1^2 + v_2^2 = 4 v_c^2 $$
$$ v_c^2 = \frac{v_1^2 + v_2^2}{4} $$