Solution
1. Geometric and Kinematic Analysis (General Variables)
Let the total length of the cord be $l$. Since the mass $m$ is suspended at the midpoint, the cord forms an isosceles triangle with the rod.
Let $\theta$ be the angle between the cord and the rod.
Coordinates:
- Horizontal position of Sleeve $M$ (relative to fixed end O): $x = l \cos \theta$
- Vertical depth of Mass $m$: $y = \frac{l}{2} \sin \theta$
Velocity Relationships:
Let $v$ be the velocity of the sleeve $M$. $$ v = \frac{dx}{dt} $$ Differentiating position $x$: $$ \frac{dx}{dt} = \frac{d}{dt}(l \cos \theta) = -l \sin \theta \cdot \frac{d\theta}{dt} $$ $$ v = -l \sin \theta \cdot \dot{\theta} \implies \dot{\theta} = \frac{-v}{l \sin \theta} $$
Now, find the velocity of mass $m$. The mass moves horizontally at half the sleeve’s speed and also moves vertically. $$ v_{m,y} = \frac{dy}{dt} = \frac{d}{dt}\left(\frac{l}{2} \sin \theta\right) = \frac{l}{2} \cos \theta \cdot \dot{\theta} $$ Substituting $\dot{\theta}$: $$ v_{m,y} = \frac{l}{2} \cos \theta \left( \frac{-v}{l \sin \theta} \right) = -\frac{v}{2} \cot \theta $$
The square of the total speed of mass $m$ ($v_m^2$) is the sum of squared horizontal and vertical components: $$ v_m^2 = (v_{m,x})^2 + (v_{m,y})^2 = \left(\frac{v}{2}\right)^2 + \left(-\frac{v}{2} \cot \theta\right)^2 $$ $$ v_m^2 = \frac{v^2}{4} (1 + \cot^2 \theta) = \frac{v^2}{4} \csc^2 \theta $$
2. Work-Energy Theorem
Applying the principle of Conservation of Energy: $$ W_{\text{ext}} + W_{\text{gravity}} = \Delta K $$
Work done by Force $F$:
Force $F$ acts horizontally on the sleeve. $$ W_F = \int F dx = F (x_{\text{final}} – x_{\text{initial}}) $$ $$ W_F = F (l \cos \theta – l \cos \theta_0) = Fl(\cos \theta – \cos \theta_0) $$
Work done by Gravity ($W_g$):
As the sleeve moves, the angle $\theta$ decreases (from $\sin^{-1}(0.8)$ to $\sin^{-1}(0.6)$), meaning the mass $m$ is pulled upwards. Gravity acts downwards, so work done is negative. $$ W_g = -mg \Delta h = -mg (y_{\text{initial}} – y_{\text{final}}) $$ $$ W_g = -mg \left( \frac{l}{2} \sin \theta_0 – \frac{l}{2} \sin \theta \right) = -\frac{mgl}{2} (\sin \theta_0 – \sin \theta) $$
Change in Kinetic Energy ($\Delta K$):
The system starts from rest, so $K_i = 0$. $$ K_f = \frac{1}{2} M v^2 + \frac{1}{2} m v_m^2 $$ Substituting $v_m^2 = \frac{v^2}{4} \csc^2 \theta$: $$ K_f = \frac{1}{2} M v^2 + \frac{1}{2} m \left( \frac{v^2}{4} \csc^2 \theta \right) = \frac{v^2}{8} (4M + m \csc^2 \theta) $$
3. General Solution for Velocity
Equating Work and Energy:
$$ Fl(\cos \theta – \cos \theta_0) – \frac{mgl}{2} (\sin \theta_0 – \sin \theta) = \frac{v^2}{8} (4M + m \csc^2 \theta) $$
Rearranging to solve for $v$:
$$ v^2 (4M + m \csc^2 \theta) = 8 \left[ Fl(\cos \theta – \cos \theta_0) – \frac{mgl}{2} (\sin \theta_0 – \sin \theta) \right] $$
$$ v^2 = \frac{4l \left[ 2F(\cos \theta – \cos \theta_0) – mg(\sin \theta_0 – \sin \theta) \right]}{4M + m \csc^2 \theta} $$
$$ v = \sqrt{ \frac{4l \{ 2F(\cos \theta – \cos \theta_0) – mg(\sin \theta_0 – \sin \theta) \}}{4M + m \csc^2 \theta} } $$
4. Numerical Substitution
Given values:
- $M = 10 \text{ kg}, \quad m = 2.0 \text{ kg}$
- $l = 102.5 \text{ cm} = 1.025 \text{ m}$
- $F = 50 \text{ N}, \quad g = 10 \text{ m/s}^2$
- Initial: $\theta_0 = \sin^{-1}(0.8) \implies \sin \theta_0 = 0.8, \cos \theta_0 = 0.6$
- Final: $\theta = \sin^{-1}(0.6) \implies \sin \theta = 0.6, \cos \theta = 0.8$
- $\csc \theta = \frac{1}{0.6} = \frac{5}{3} \implies \csc^2 \theta = \frac{25}{9}$
Calculating Numerator Terms:
- $2F(\cos \theta – \cos \theta_0) = 2(50)(0.8 – 0.6) = 100(0.2) = 20$
- $mg(\sin \theta_0 – \sin \theta) = 2(10)(0.8 – 0.6) = 20(0.2) = 4$
- Term in brackets: $\{ 20 – 4 \} = 16$
- Total Numerator: $4l \times 16 = 4(1.025)(16) = 65.6$
Calculating Denominator Terms:
- $4M + m \csc^2 \theta = 4(10) + 2\left(\frac{25}{9}\right)$
- $= 40 + \frac{50}{9} = \frac{360 + 50}{9} = \frac{410}{9}$
Final Calculation:
$$ v^2 = \frac{65.6}{\frac{410}{9}} = \frac{65.6 \times 9}{410} $$
$$ v^2 = \frac{590.4}{410} = 1.44 $$
$$ v = \sqrt{1.44} = 1.2 \text{ m/s} $$
Answer: 1.2 m/s