NLM BYU 25

1. Physics Principle: Theorem of Equal Times
The problem states that the bead takes equal time to slide down chords $AB$ and $AC$ starting from rest at $A$. For a particle sliding down a smooth chord under gravity starting from the highest point, the time of descent is constant for all chords ending on the circle.

Therefore, points $A$, $B$, and $C$ must lie on a vertical circle with $A$ at the highest point (the vertex). The tangent to the circle at $A$ is horizontal.

2. Geometric Properties of Chords
We use the “Alternate Segment Theorem” or the property of tangent-chord angles: The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment.

• Angle between horizontal tangent at $A$ and chord $AB$ = Angle subtended by $AB$ at $C$ = $\angle C = 55^\circ$.
• Angle between horizontal tangent at $A$ and chord $AC$ = Angle subtended by $AC$ at $B$ = $\angle B = 74^\circ$.

This means chord $AB$ makes an angle of $55^\circ$ with the horizontal, and chord $AC$ makes an angle of $74^\circ$ with the horizontal.

3. Calculating the Angle $\theta$
We need to find the angle $\theta$ that the arm $BC$ makes with the horizontal. Let’s analyze the angles at point $B$.

• The vector $\vec{BA}$ (directed upwards) makes an angle of $55^\circ$ with the horizontal.
• The given angle between rods at corner $B$ is $\angle ABC = 74^\circ$.

Since $C$ is further down the circle (steeper chord from A), the arm $BC$ must point downwards relative to the horizontal line passing through $B$. Using vector subtraction of angles or simple geometry at $B$: $$ \text{Angle}(\vec{BC}, \text{Horizontal}) = |\text{Angle}(\vec{BA}, \text{Horizontal}) – \angle ABC| $$ $$ \theta = |55^\circ – 74^\circ| $$ $$ \theta = |-19^\circ| = 19^\circ $$

Answer:
The angle $\theta$ which the arm $BC$ makes with the horizontal is $19^\circ$.