Use the below information to answer Questions 12 – 13. A uniform bar, of mass M, with seven evenly spaced holes is held by sliding the bar over a horizontal peg through one of the seven holes. The peg passes through hole D, and a cylinder hangs from a hook placed through hole G as shown below. The mass of the bar is the same as the mass of the cylinder, and the location of the center of mass of the bar is at the center of hole D. In this configuration, the bar-cylinder system is not in rotational equilibrium and is free to rotate about the peg in hole D. Frictional forces acting on the bar are negligible. A в сD E F G оо (12) Two additional cylinders, identical to the one shown, can each be hung on the bar by hooks from any of the open holes so that the bar will be in rotational equilibrium when all three cylinders are hanging from it. From which two holes should the two cylinders be hung? Briefly explain your reasoning. The bar is returned to its original setup, Case 1, in which the cylinder has weight F, as shown below. The bar is released, and its resulting angular acceleration is recorded. Then, in Case 2, the weight is replaced by a string, as shown below. A downward force of magnitude also equal to F is exerted on the string. The resulting angular acceleration of the bar is recorded. Case 1 Weight = F Case 2 Force = F
Use the below information to answer Questions 12 – 13. A uniform bar, of mass M, with seven evenly spaced holes is held by sliding the bar over a horizontal peg through one of the seven holes. The peg passes through hole D, and a cylinder hangs from a hook placed through hole G as shown below. The mass of the bar is the same as the mass of the cylinder, and the location of the center of mass of the bar is at the center of hole D. In this configuration, the bar-cylinder system is not in rotational equilibrium and is free to rotate about the peg in hole D. Frictional forces acting on the bar are negligible. A в сD E F G оо (12) Two additional cylinders, identical to the one shown, can each be hung on the bar by hooks from any of the open holes so that the bar will be in rotational equilibrium when all three cylinders are hanging from it. From which two holes should the two cylinders be hung? Briefly explain your reasoning. The bar is returned to its original setup, Case 1, in which the cylinder has weight F, as shown below. The bar is released, and its resulting angular acceleration is recorded. Then, in Case 2, the weight is replaced by a string, as shown below. A downward force of magnitude also equal to F is exerted on the string. The resulting angular acceleration of the bar is recorded. Case 1 Weight = F Case 2 Force = F
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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