Solution to Problem 2
Problem Statement: A bead moves in a circle of radius $r$ on a floor with speed $v_0$. The thread passes through a hole/center and is pulled up with acceleration $a_0$. We need to find the acceleration of the bead immediately after the pull begins.
1. Geometric Setup:
Let the length of the string be $L$. The string makes an angle $\theta$ with the vertical. The radius of the circular path on the floor is $r$.
$$ r = l \sin \theta $$
Since the bead is on the floor, its motion is horizontal.
2. Acceleration Components:
The bead experiences two components of acceleration in the horizontal plane:
- Centripetal Acceleration ($a_c$): Due to the circular motion with speed $v_0$. $$ a_c = \frac{v_0^2}{r} = \frac{v_0^2}{l \sin \theta} $$ This is directed radially inward.
- Radial “Pull” Acceleration ($a_r$): This arises because the string is being accelerated upwards. We must use the constraint relation between the vertical motion of the hand and the horizontal motion of the bead.
3. Constraint Analysis:
Let $y$ be the vertical length of the string above the floor and $r$ be the radial distance on the floor. The total length of the string involved is $l = \sqrt{r^2 + y^2}$.
Differentiating with respect to time (assuming the string is taut):
$$ \dot{l} = \frac{r\dot{r} + y\dot{y}}{l} $$
Since the string is inextensible and we are looking at the acceleration transmission, we consider the component of accelerations along the string.
Alternatively, consider the velocity components along the string. $$ v_{string} = -v_{bead\_radial} \sin \theta + v_{hand} \cos \theta $$ Assuming the string length is constant (between hand and bead), the relative velocity along the line is zero. $$ v_{bead\_radial} \sin \theta = v_{hand} \cos \theta $$ Differentiating for acceleration (at the instant where velocities might be zero or constant, but accelerations are $a_0$ and $a_r$): $$ a_{bead\_radial} \sin \theta = a_{hand} \cos \theta $$ Given $a_{hand} = a_0$ (upwards), $$ a_{bead\_radial} = a_0 \frac{\cos \theta}{\sin \theta} = a_0 \cot \theta $$ This acceleration is also directed radially inward (towards the center).
4. Total Acceleration:
Both the centripetal acceleration required to turn the particle and the radial acceleration due to the shortening radius are directed towards the center of the circle.
$$ a_{total} = a_c + a_{bead\_radial} $$
$$ a_{total} = \frac{v_0^2}{l \sin \theta} + a_0 \cot \theta $$
Answer: $$ \frac{v_0^2}{l \sin \theta} + a_0 \cot \theta \quad \text{(towards the centre)} $$