Ch. 09 - Linear Momentum

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 09 - Linear Momentum

Problem 44

Chapter 9, Problem 44

(II) A pendulum consists of a mass M hanging at the bottom end of a massless rod of length ℓ, which has a frictionless pivot at its top end. A mass m, moving horizontally as shown in Fig. 9–44 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc?

Verified step by step guidance

Step 1: Analyze the problem and identify the key concepts. This problem involves conservation of momentum during the collision and conservation of energy for the pendulum's motion. The goal is to find the minimum velocity \(v\) such that the pendulum swings over the top of its arc.

Step 2: Apply the principle of conservation of linear momentum for the collision. Before the collision, the moving mass \(m\) has momentum \(p = mv\), and the pendulum mass \(M\) is stationary. After the collision, the combined system (\(m + M\)) moves with a velocity \(v'\). The equation is: \[ mv = (m + M)v' \]. Solve for \(v'\): \[ v' = \frac{mv}{m + M} \].

Step 3: Determine the energy required for the pendulum to swing over the top. For the pendulum to just clear the top, its center of mass must reach a height of \(2\ell\) (the length of the rod). The total mechanical energy at the bottom (kinetic energy) must equal the potential energy at the top. The potential energy at the top is \(U = (m + M)g(2\ell)\).

Step 4: Relate the kinetic energy of the pendulum after the collision to its velocity \(v'\). The kinetic energy immediately after the collision is \(K = \frac{1}{2}(m + M)v'^2\). Set this equal to the potential energy at the top: \[ \frac{1}{2}(m + M)v'^2 = (m + M)g(2\ell) \]. Simplify to find \(v'\): \[ v'^2 = 4g\ell \].

Step 5: Combine the results from Step 2 and Step 4 to find the minimum \(v\). Substitute \(v' = \sqrt{4g\ell}\) into \(v' = \frac{mv}{m + M}\): \[ \frac{mv}{m + M} = \sqrt{4g\ell} \]. Solve for \(v\): \[ v = \frac{(m + M)}{m} \sqrt{4g\ell} \]. This is the minimum velocity required for the pendulum to swing over the top.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In the context of the pendulum question, when mass m collides with mass M, the momentum before the collision must equal the momentum after the collision, allowing us to calculate the necessary velocity v for mass m.

Energy Conservation

Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the kinetic energy of the combined masses after the collision must be sufficient to convert into gravitational potential energy to lift the pendulum over the top of its arc, which is crucial for determining the minimum velocity v.

Pendulum Motion

Pendulum motion refers to the oscillatory movement of a mass attached to a rod swinging about a pivot. The maximum height reached by the pendulum is determined by the initial kinetic energy imparted to it. Understanding the dynamics of pendulum motion is essential for calculating the conditions under which the pendulum will swing over the top of its arc after the collision.

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