Solution to Question 40

1. Kinematic Relations:
Let $\theta$ be the angular position of the line connecting center O, point B, and point A, measured from the vertical (radius to the tangent point).
From the geometry, the position of A on the straight track is: $$ x_A = r \tan\theta $$ Differentiating with respect to time to relate velocities ($v_A = u$): $$ u = \frac{dx_A}{dt} = r \sec^2\theta \cdot \frac{d\theta}{dt} $$ Thus, the angular velocity of the line OBA is: $$ \omega = \frac{d\theta}{dt} = \frac{u}{r} \cos^2\theta $$

2. Motion of Point B:
Point B moves on the circular track of radius $r$. Its velocity is purely tangential. $$ v_B = r\omega = r \left( \frac{u}{r} \cos^2\theta \right) = u \cos^2\theta $$ To find the total acceleration of B, we need the tangential component ($a_t$) and the normal (centripetal) component ($a_n$).

3. Acceleration Components:
Tangential Acceleration ($a_t$): $$ a_t = \frac{dv_B}{dt} = \frac{d}{dt}(u \cos^2\theta) = u \cdot 2\cos\theta (-\sin\theta) \cdot \frac{d\theta}{dt} $$ Substitute $\frac{d\theta}{dt} = \frac{u}{r} \cos^2\theta$: $$ a_t = -u (2\sin\theta \cos\theta) \left( \frac{u}{r} \cos^2\theta \right) = -\frac{2u^2}{r} \sin\theta \cos^3\theta $$ Normal Acceleration ($a_n$): $$ a_n = \frac{v_B^2}{r} = \frac{(u \cos^2\theta)^2}{r} = \frac{u^2}{r} \cos^4\theta $$

4. Total Acceleration:
The magnitude of acceleration $a_B$ is $\sqrt{a_t^2 + a_n^2}$. $$ a_B = \sqrt{ \left( -\frac{2u^2}{r} \sin\theta \cos^3\theta \right)^2 + \left( \frac{u^2}{r} \cos^4\theta \right)^2 } $$ Factor out common terms $\frac{u^2}{r} \cos^3\theta$: $$ a_B = \frac{u^2}{r} \cos^3\theta \sqrt{ (-2\sin\theta)^2 + (\cos\theta)^2 } $$ $$ a_B = \frac{u^2}{r} \cos^3\theta \sqrt{ 4\sin^2\theta + \cos^2\theta } $$ Using $\cos^2\theta = 1 – \sin^2\theta$: $$ 4\sin^2\theta + (1 – \sin^2\theta) = 1 + 3\sin^2\theta $$ $$ a_B = \frac{u^2}{r} \cos^3\theta \sqrt{ 1 + 3\sin^2\theta } $$