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A 1.80-kg monkey wrench is pivoted 0.250 m from its center

Problem 54E Chapter 14

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 54E

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

Step-by-Step Solution:

Solution 54E Problem (a) Step 1: Mass of the wrench m = 1.80 kg Distance from center of gravity d = 0.25 m Time period of oscillation T = 0.94 s Step 2: To find the moment of inertia of the wrench about an axis through the pivot Time period for the oscillations in physical pendulum is I T = 2 mgd ----(1) Where g is acceleration due to gravity (9.8 m/s ) 2 Step 3: Rearranging the equation (1) mgdT2 I = 42 -----(2) Substituting known numerical values in (2) 2 I = 1.* 9*8 0.25 0.94 4*3.1415 I = 0.0987 Kg.m 2 The moment of inertia of the wrench about an axis through the pivot is 0.0987 2 Kg.m Problem (b) Step 1: Angular displacement of the wrench from its initial position = 0.40 rad To find the angular speed of wrench The wrench is displaced 0.40 rad from its equilibrium position. Potential energy at the new displaced position P h2 to be found. The diagram shows the change in angular displacement of the wrench about its mean position. At P the potential energy of the wrench is 2 U = mgd(1-cos) Step 2: When we release the wrench the potential energy is converted into kinetic energy. Therefore mgd(1-cos) = I -----(3) 2 Where is angular speed of the wrench. From equation (2) I.4 mgd = T2 ----- (4)

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