Balancing Act. Attached to one end of a long, thin,

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

Problem 66P Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 66P

Balancing Act. ?Attached to one end of a long, thin, uniform rod of length ?L ?and mass ?M ?is a small blob of clay of the same mass ?M. ?(a) Locate the position of the center of mass of the system of rod and clay. Note this position on a drawing of the rod. (b) You carefully balance the rod on a frictionless tabletop so that it is standing vertically, with the end without the clay touching the table. If the rod is now tipped so that it is a small angle ? away from the vertical, determine its angular acceleration at this instant. Assume that the end without the clay remains in contact with the tabletop. (?Hint: ?See Table 9.2.) (c) You again balance the rod on the frictionless tabletop so that it is standing vertically, but now the end of the rod ?with ?the clay is touching the table. If the rod is again tipped so that it is a small angle ? away from the vertical, determine its angular acceleration at this instant. Assume that the end with the clay remains in contact with the tabletop. How does this compare to the angular acceleration in part (b)? (d) A pool cue is a tapered wooden rod that is thick at one end and thin at the other. You can easily balance a pool cue vertically on one finger if the thin end is in contact with your finger; this is quite a bit harder to do if the thick end is in contact with your finger. Explain why there

Step-by-Step Solution:

Solution 66P Step 1: The mass of the rod is M, Length of rod=L The mass of the clay is M clay Step 2: (a).From drawing below, we get that center of mass of rod is C and center of mass of clay at Point B.so the center of mass of the system is at L/4 from point B. Step 2: (b). Moment of inertia is given by I = 4/3ML 2 Torque is find out by = (3L/4)(2 mg sin) = (3 mgL/2) sin) Torque is defined as = I/ = /I = (3 mgL/2) sin) /(4/3)ML 2 = (9g/8L) sin

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Chapter 10, Problem 66P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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