Solution to the January, 2018 Challenge Touched by an angle*

Solution to the January, 2018 Challenge Touched by an angle* Physics Challenge for Teachers and Students Method 1 The figure shows the free-body diagram of the system. The light horizontal rod, for small- amplitude oscillations, move a distance x to the right from its equilibrium position. x B C  l Mg Let us call F and F the forces in the pivots B and C. 1 2 There is no acceleration in the vertical direction, so the equations of motion for the sphere in the vertical and horizontal direction are, respectively 𝐹 𝑜𝑠𝑐𝜃 + 𝐹 𝑜𝑠𝑐𝜃 − = 0 (1) 1 2 −𝐹 𝑖𝑛𝜃𝑠 + 𝐹 𝜃𝑠𝑖𝑛 = 𝑀 𝑥 ̈ (2) 1 2 For small-amplitude oscillations 𝑜𝑠𝑐𝜃 ≈ 1 and (1) is 𝐹 + 𝐹 = (3) 1 2 The equation for rotational equilibrium relative to the midpoint of the rod is 𝐿 𝐿 −𝐹 𝜃𝑜𝑠𝑐 + 𝐹 𝑜𝑠𝑐𝜃 + = 0 (4) 1 2 2 2 for small-amplitude oscillations we obtain for (4) 𝑀𝑔𝑑 𝐹 − 𝐹 = −2 (5) 2 1 From equations (3) and (5) we obtain 𝑀𝑔 2𝑑 𝐹 = (1 + ) (6) 2 𝐿 𝑀𝑔 2𝑑 𝐹 = (1 − ) (7) 2 𝐿 For small-amplitude oscillations 𝑠𝑖𝑛𝜃 ≈ 𝜃 and http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png The Physics Teacher American Association of Physics Teachers

Solution to the January, 2018 Challenge Touched by an angle*

The Physics Teacher , Volume 56 (3): 1 – Mar 1, 2018

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Publisher
American Association of Physics Teachers
Copyright
© American Association of Physicists in Medicine.
ISSN
0031-921X
D.O.I.
10.1119/1.5025279
Publisher site
See Article on Publisher Site

Abstract

Physics Challenge for Teachers and Students Method 1 The figure shows the free-body diagram of the system. The light horizontal rod, for small- amplitude oscillations, move a distance x to the right from its equilibrium position. x B C  l Mg Let us call F and F the forces in the pivots B and C. 1 2 There is no acceleration in the vertical direction, so the equations of motion for the sphere in the vertical and horizontal direction are, respectively 𝐹 𝑜𝑠𝑐𝜃 + 𝐹 𝑜𝑠𝑐𝜃 − = 0 (1) 1 2 −𝐹 𝑖𝑛𝜃𝑠 + 𝐹 𝜃𝑠𝑖𝑛 = 𝑀 𝑥 ̈ (2) 1 2 For small-amplitude oscillations 𝑜𝑠𝑐𝜃 ≈ 1 and (1) is 𝐹 + 𝐹 = (3) 1 2 The equation for rotational equilibrium relative to the midpoint of the rod is 𝐿 𝐿 −𝐹 𝜃𝑜𝑠𝑐 + 𝐹 𝑜𝑠𝑐𝜃 + = 0 (4) 1 2 2 2 for small-amplitude oscillations we obtain for (4) 𝑀𝑔𝑑 𝐹 − 𝐹 = −2 (5) 2 1 From equations (3) and (5) we obtain 𝑀𝑔 2𝑑 𝐹 = (1 + ) (6) 2 𝐿 𝑀𝑔 2𝑑 𝐹 = (1 − ) (7) 2 𝐿 For small-amplitude oscillations 𝑠𝑖𝑛𝜃 ≈ 𝜃 and

Journal

The Physics TeacherAmerican Association of Physics Teachers

Published: Mar 1, 2018

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