Question

Find tension in the thread when the extreme discs are about to collide.

Answer 1

4 N

Answer 2

1. System Description: The system consists of a central mass $M$ and two identical outer masses $m$, connected by threads of length $l$. The central mass $M$ is given an initial velocity $u$ perpendicular to the threads on a smooth horizontal table. We need to find the tension in the threads when the outer masses are about to collide.

2. Coordinate System and Constraints: Let the initial position of the central mass $M$ be at the origin $(0,0)$. The initial positions of the outer masses are $(-l, 0)$ and $(l, 0)$. Let the positions at time $t$ be $M(0, y_M)$, $m_1(-x, y)$, and $m_2(x, y)$. The constraint of constant thread length gives: $x^2 + (y_M - y)^2 = l^2$. The outer masses collide when $x=0$.

3. Conservation of Momentum: In the y-direction, the initial momentum is $M u$. At any time $t$, the y-momentum is $M v_{My} + m v_{m_1y} + m v_{m_2y}$. Since $v_{m_1y} = v_{m_2y} = \dot{y}$ and $v_{My} = \dot{y}_M$, we have: $M u = M \dot{y}_M + 2m \dot{y}$.

4. Conservation of Energy: The initial kinetic energy is $\frac{1}{2} M u^2$. At time $t$, the kinetic energy is $\frac{1}{2} M v_M^2 + \frac{1}{2} m v_{m_1}^2 + \frac{1}{2} m v_{m_2}^2$. Since the motion is along the y-axis at the initial moment and later, $v_M^2 = \dot{y}_M^2$. The velocities of the outer masses are $v_{m_1}^2 = (-\dot{x})^2 + \dot{y}^2 = \dot{x}^2 + \dot{y}^2$ and $v_{m_2}^2 = \dot{x}^2 + \dot{y}^2$. So, $\frac{1}{2} M u^2 = \frac{1}{2} M \dot{y}_M^2 + 2 \times \frac{1}{2} m (\dot{x}^2 + \dot{y}^2)$, which simplifies to $M u^2 = M \dot{y}_M^2 + 2m (\dot{x}^2 + \dot{y}^2)$.

5. Constraint Differentiation: Differentiating $x^2 + (y_M - y)^2 = l^2$ with respect to time: $2x \dot{x} + 2(y_M - y)(\dot{y}_M - \dot{y}) = 0 \implies x \dot{x} + (y_M - y)(\dot{y}_M - \dot{y}) = 0$.

6. Condition at Collision: Just before collision, $x=0$. From the constraint equation, $(y_M - y)^2 = l^2$, so $|y_M - y| = l$. Since the initial velocity is in the positive y-direction, $y_M > y$, thus $y_M - y = l \neq 0$. The differentiated constraint equation at $x=0$ becomes $(y_M - y)(\dot{y}_M - \dot{y}) = 0$, which implies $\dot{y}_M - \dot{y} = 0$, so $\dot{y}_M = \dot{y}$.

7. Velocities at Collision: Substitute $\dot{y}_M = \dot{y}$ into the momentum equation: $M u = M \dot{y} + 2m \dot{y} = (M + 2m) \dot{y} \implies \dot{y} = \dot{y}_M = \frac{M u}{M + 2m}$. Substitute $\dot{y}_M = \dot{y}$ into the energy equation: $M u^2 = M \dot{y}^2 + 2m (\dot{x}^2 + \dot{y}^2) = (M + 2m) \dot{y}^2 + 2m \dot{x}^2$. $M u^2 = (M + 2m) \left(\frac{M u}{M + 2m}\right)^2 + 2m \dot{x}^2$. $M u^2 = \frac{M^2 u^2}{M + 2m} + 2m \dot{x}^2$. $2m \dot{x}^2 = M u^2 - \frac{M^2 u^2}{M + 2m} = M u^2 \left(\frac{M + 2m - M}{M + 2m}\right) = \frac{2m M u^2}{M + 2m}$. $\dot{x}^2 = \frac{M u^2}{M + 2m}$.

8. Tension Calculation: Consider the motion of an outer mass $m$ relative to the central mass $M$. Just before collision, $x=0$, so the thread is along the y-axis. The relative velocity of $m$ with respect to $M$ is $\vec{v}_{m/M} = \vec{v}_m - \vec{v}_M = (-\dot{x}, \dot{y}) - (0, \dot{y}_M) = (-\dot{x}, 0)$ since $\dot{y} = \dot{y}_M$. The tangential velocity magnitude is $v_{tan} = |\dot{x}| = \sqrt{\frac{M u^2}{M + 2m}}$. The tension $T$ provides the centripetal force for this relative motion: $T = m \frac{v_{tan}^2}{l} = m \frac{\dot{x}^2}{l} = m \frac{1}{l} \left(\frac{M u^2}{M + 2m}\right)$.

9. Substituting Given Values: $u = 2$ m/s, $M = 2$ kg, $m = 1$ kg, $l = 0.5$ m. $T = (1 \text{ kg}) \times \frac{1}{0.5 \text{ m}} \times \frac{(2 \text{ kg}) \times (2 \text{ m/s})^2}{2 \text{ kg} + 2 \times (1 \text{ kg})}$. $T = 2 \times \frac{2 \times 4}{2 + 2} = 2 \times \frac{8}{4} = 2 \times 2 = 4$ N.