Question

A block of mass $M$ can slide without friction on a horizontal floor. A small ball of mass $m$ is suspended by a light inextensible cord of length $l$ from a hanger fixed at the top of a vertical pole mounted on the block so that the ball can swing freely in a vertical plane. The ball is released from a horizontal position as shown. Find the maximum tensile force in the string during the subsequent motion.

• A. The maximum tensile force in the string is $mg \left(3 + \frac{2m}{M}\right)$.
• B. The maximum tensile force in the string is $mg \left(1 + \frac{2m}{M}\right)$.
• C. The maximum tensile force in the string is $mg \left(2 + \frac{2m}{M}\right)$.
• D. The maximum tensile force in the string is $mg \left(3 + \frac{m}{M}\right)$.

Answer 1

Answer: (A)

The maximum tensile force in the string is $mg \left(3 + \frac{2m}{M}\right)$.

Answer 2

The problem involves analyzing the motion of a pendulum attached to a movable block. We use the principles of conservation of horizontal momentum and conservation of mechanical energy to relate the velocities and positions of the ball and the block.

1. Conservation of Horizontal Momentum: Since there are no external horizontal forces on the system, the total horizontal momentum is conserved. Let $v_M$ be the velocity of the block and $v_{m,x}$ be the horizontal velocity of the ball. Initially, both are zero. At any point, $M v_M + m v_{m,x} = 0$.

2. Conservation of Energy: The total mechanical energy is conserved. We set the initial potential energy to zero. At an angle $\theta$ with the vertical, the potential energy is $PE = -mgl \cos\theta$. The kinetic energy is $KE = \frac{1}{2}M v_M^2 + \frac{1}{2}m v_m^2$. Thus, the total energy is $E = \frac{1}{2}M v_M^2 + \frac{1}{2}m v_m^2 - mgl \cos\theta = 0$.

3. Relating Velocities: Let $\theta$ be the angle with the vertical and $\dot{\theta}$ be the angular velocity of the pendulum relative to the block. The absolute velocity of the ball $\vec{v}_m$ is the vector sum of the block's velocity $\vec{v}_M$ and the ball's velocity relative to the suspension point $\vec{v}_{m/susp}$. $\vec{v}_m = \vec{v}_M + \vec{v}_{m/susp}$. From momentum conservation, $v_M = -\frac{m}{M}v_{m,x}$. By considering the velocity components and using momentum conservation, we find $v_M = -\frac{ml\dot{\theta}\cos\theta}{M+m}$. The square of the ball's absolute speed is $v_m^2 = l^2\dot{\theta}^2 \left(\frac{M^2\cos^2\theta}{(M+m)^2} + \sin^2\theta\right)$.

4. Energy Equation in terms of $\dot{\theta}$: Substituting $v_M$ and $v_m^2$ into the energy conservation equation and solving for $\dot{\theta}^2$, we get: $\dot{\theta}^2 = \frac{2g\cos\theta}{l \left(\frac{M\cos^2\theta}{M+m} + \sin^2\theta\right)}$.

5. Tensile Force: The tensile force $T$ in the string is given by Newton's second law in the radial direction: $T - mg\cos\theta = m(l\dot{\theta}^2)$ (where the acceleration is $l\dot{\theta}^2$ towards the center of the circular path). $T = mg\cos\theta + ml\dot{\theta}^2$.

6. Maximum Tensile Force: Substituting the expression for $\dot{\theta}^2$: $T = mg\cos\theta + ml \left(\frac{2g\cos\theta}{l \left(\frac{M\cos^2\theta}{M+m} + \sin^2\theta\right)}\right) = mg\cos\theta + \frac{2mg\cos\theta}{\frac{M\cos^2\theta}{M+m} + \sin^2\theta}$. To maximize $T$, we analyze the expression as a function of $\cos\theta$. Let $u = \cos\theta$. $T(u) = mgu + \frac{2mgu}{\frac{Mu^2}{M+m} + (1-u^2)}$. The derivative of $T(u)$ with respect to $u$ is positive for $u \in [0, 1]$. Thus, $T$ is maximum when $\cos\theta$ is maximum, which occurs at $\theta = 0$ (the lowest point).

7. Calculation at $\theta=0$: At $\theta = 0$, $\cos\theta = 1$. $T_{max} = mg(1) + \frac{2mg(1)}{\frac{M(1)^2}{M+m} + (0)^2} = mg + \frac{2mg}{\frac{M}{M+m}} = mg + \frac{2mg(M+m)}{M}$. $T_{max} = mg + 2mg \left(1 + \frac{m}{M}\right) = mg + 2mg + \frac{2m^2g}{M} = 3mg + \frac{2m^2g}{M}$. $T_{max} = mg \left(3 + \frac{2m}{M}\right)$.