Problem 1.55 — Velocity of rope end on a pulley

Problem Statement

Solve the Newton’s Laws / mechanics problem: A rope passes over a fixed pulley of radius $R$ at the top of a vertical wall. One end hangs vertically, the other goes horizontally and is pulled at speed $v_0$. Find the speed at which the hanging end descends. The rope is inextensible. If one end is pulled at $v_0$ (horizontally), the linear spee

Given Information

  • See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

  • See the step-by-step solution for the specific equations applied.
  • All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$

$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$

$$T = \frac{2m_1 m_2 g}{m_1+m_2} = \frac{2\times3\times5\times9.8}{8} = \frac{294}{8} = 36.75\,\text{N}$$

Answer

$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.


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