Question

A sleeve of mass $M = 10$ kg can slide on a horizontal frictionless rod. A block of mass $m = 2.0$ kg is suspended at the mid-point of a light inextensible cord of length $l = 102.5$ cm, one end of which is attached to the sleeve and the other to one end of the rod. Initially the sleeve is held at rest and the angle between the rod and a cord segment is $\theta = \theta_0 = \sin^{-1}(0.8)$ as shown in the figure. If a constant horizontal force $F = 50$ N is applied on the sleeve, find its speed when the angle between cord segments and the rod becomes $\theta = \sin^{-1}(0.6)$?

Answer 1

1.2 m/s

Answer 2

The problem involves a system of a sleeve and a block connected by a cord, moving under the influence of an external horizontal force and gravity. We can use the work-energy theorem for the system to find the final speed of the sleeve.

1. Define System and Coordinate System:

The system consists of the sleeve (mass $M$) and the block (mass $m$). The rod is horizontal and frictionless. Let the fixed end of the rod be the origin (0,0). The sleeve moves along the x-axis. The block moves in the x-y plane.

2. Express Positions in terms of Angle $\theta$:

The total length of the cord is $l$. Since the block is suspended at the mid-point, each segment of the cord has length $l/2$. Let the angle between the rod and each cord segment be $\theta$. The vertical depth of the block below the rod, $h$, is given by: $h = \frac{l}{2} \sin \theta$ The horizontal position of the block, $x_m$, from the fixed end is: $x_m = \frac{l}{2} \cos \theta$ The horizontal position of the sleeve, $x_M$, from the fixed end is the sum of the horizontal projections of both cord segments: $x_M = \frac{l}{2} \cos \theta + \frac{l}{2} \cos \theta = l \cos \theta$

3. Calculate Work Done by External Force ($W_F$):

The constant horizontal force $F$ acts on the sleeve. The work done by $F$ is $W_F = F \Delta x_M$. Initial angle $\theta_0 = \sin^{-1}(0.8)$, so $\sin \theta_0 = 0.8$. This implies $\cos \theta_0 = \sqrt{1 - (0.8)^2} = 0.6$. Final angle $\theta = \sin^{-1}(0.6)$, so $\sin \theta = 0.6$. This implies $\cos \theta = \sqrt{1 - (0.6)^2} = 0.8$. Initial position of sleeve: $x_{M,i} = l \cos \theta_0$. Final position of sleeve: $x_{M,f} = l \cos \theta$. $W_F = F (x_{M,f} - x_{M,i}) = F (l \cos \theta - l \cos \theta_0)$ Given $F = 50$ N, $l = 102.5$ cm = 1.025 m. $W_F = 50 \times 1.025 \times (0.8 - 0.6) = 50 \times 1.025 \times 0.2 = 10.25$ J.

4. Calculate Change in Potential Energy ($\Delta U$):

Only the gravitational potential energy of the block changes. Let the rod level be the reference ($U=0$). Initial potential energy of block: $U_i = -m g h_i = -m g \left(\frac{l}{2} \sin \theta_0\right)$. Final potential energy of block: $U_f = -m g h_f = -m g \left(\frac{l}{2} \sin \theta\right)$. Change in potential energy: $\Delta U = U_f - U_i = -\frac{1}{2} m g l (\sin \theta - \sin \theta_0) = \frac{1}{2} m g l (\sin \theta_0 - \sin \theta)$. Given $m = 2.0$ kg, $g = 10$ m/s$^2$ (standard assumption for clean numbers unless specified otherwise). $\Delta U = \frac{1}{2} \times 2.0 \times 10 \times 1.025 \times (0.8 - 0.6) = 10 \times 1.025 \times 0.2 = 2.05$ J. (The potential energy increases because the block moves upwards as $\theta$ decreases).

5. Relate Velocities of Sleeve and Block:

Let $V_M$ be the speed of the sleeve and $v$ be the speed of the block. Differentiating $x_M = l \cos \theta$ with respect to time: $V_M = \left| \frac{dx_M}{dt} \right| = \left| -l \sin \theta \frac{d\theta}{dt} \right| = l \sin \theta \left| \frac{d\theta}{dt} \right|$. The position vector of the block is $\vec{r}_m = \left(\frac{l}{2} \cos \theta\right) \hat{i} - \left(\frac{l}{2} \sin \theta\right) \hat{j}$. Differentiating with respect to time: $\vec{v} = \frac{d\vec{r}_m}{dt} = -\frac{l}{2} \sin \theta \frac{d\theta}{dt} \hat{i} - \frac{l}{2} \cos \theta \frac{d\theta}{dt} \hat{j}$. The speed of the block is $v = |\vec{v}| = \sqrt{\left(-\frac{l}{2} \sin \theta \frac{d\theta}{dt}\right)^2 + \left(-\frac{l}{2} \cos \theta \frac{d\theta}{dt}\right)^2}$ $v = \sqrt{\left(\frac{l}{2}\right)^2 \left(\frac{d\theta}{dt}\right)^2 (\sin^2 \theta + \cos^2 \theta)} = \frac{l}{2} \left| \frac{d\theta}{dt} \right|$. From $V_M = l \sin \theta \left| \frac{d\theta}{dt} \right|$, we get $\left| \frac{d\theta}{dt} \right| = \frac{V_M}{l \sin \theta}$. Substitute this into the expression for $v$: $v = \frac{l}{2} \left(\frac{V_M}{l \sin \theta}\right) = \frac{V_M}{2 \sin \theta}$.

6. Calculate Change in Kinetic Energy ($\Delta K$):

Initial kinetic energy $K_i = 0$ (sleeve and block are initially at rest). Final kinetic energy $K_f = \frac{1}{2} M V_M^2 + \frac{1}{2} m v^2$. Substitute $v = \frac{V_M}{2 \sin \theta}$: $K_f = \frac{1}{2} M V_M^2 + \frac{1}{2} m \left(\frac{V_M}{2 \sin \theta}\right)^2 = \frac{1}{2} V_M^2 \left(M + \frac{m}{4 \sin^2 \theta}\right)$. $\Delta K = K_f - K_i = K_f$.

7. Apply Work-Energy Theorem:

$W_{ext} = \Delta K + \Delta U$ $W_F = \frac{1}{2} V_M^2 \left(M + \frac{m}{4 \sin^2 \theta}\right) + \Delta U$ $10.25 = \frac{1}{2} V_M^2 \left(10 + \frac{2}{4 \times (0.6)^2}\right) + 2.05$ $10.25 - 2.05 = \frac{1}{2} V_M^2 \left(10 + \frac{2}{4 \times 0.36}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(10 + \frac{1}{2 \times 0.36}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(10 + \frac{1}{0.72}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(10 + \frac{100}{72}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(10 + \frac{25}{18}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(\frac{180 + 25}{18}\right)$ $8.2 = \frac{1}{2} V_M^2 \left(\frac{205}{18}\right)$ $16.4 = V_M^2 \left(\frac{205}{18}\right)$ $V_M^2 = \frac{16.4 \times 18}{205} = \frac{295.2}{205} = 1.44$ $V_M = \sqrt{1.44} = 1.2$ m/s.

The speed of the sleeve when the angle becomes $\sin^{-1}(0.6)$ is 1.2 m/s.

Explanation of the solution:

1. Identify Initial and Final States: Determine given parameters ($M, m, l, F, \theta_0, \theta$) for initial (rest) and final states.
2. Express Positions: Relate the sleeve's horizontal position ($x_M$) and the block's vertical position ($h$) to the angle $\theta$ using trigonometry and the cord length $l$. Specifically, $x_M = l \cos \theta$ and $h = (l/2) \sin \theta$.
3. Work-Energy Theorem: Apply the work-energy theorem for the system (sleeve + block): $W_{ext} = \Delta K + \Delta U$.
4. Calculate Work Done by External Force: $W_F = F \cdot \Delta x_M = F (l \cos \theta - l \cos \theta_0)$.
5. Calculate Change in Potential Energy: $\Delta U = m g (h_i - h_f) = \frac{1}{2} m g l (\sin \theta_0 - \sin \theta)$.
6. Relate Velocities: Differentiate the position equations with respect to time to find the speeds of the sleeve ($V_M$) and block ($v$). Establish the relationship $v = \frac{V_M}{2 \sin \theta}$.
7. Calculate Change in Kinetic Energy: $\Delta K = K_f - K_i = \frac{1}{2} M V_M^2 + \frac{1}{2} m v^2$. Substitute the velocity relation to express $\Delta K$ solely in terms of $V_M$.
8. Solve for $V_M$: Substitute all calculated values into the work-energy equation and solve for $V_M$.

Answer:

The speed of the sleeve when the angle between cord segments and the rod becomes $\theta = \sin^{-1}(0.6)$ is 1.2 m/s.