Angular Momentum Sum: Calculating Work Done

AI Thread Summary
The discussion focuses on calculating the work done by a force acting on a horizontal rod pivoted at one end. The rod rotates about a vertical axis under the influence of a steady force applied at its free end. To determine the work done, one must consider the distance over which the force acts during one complete turn. The force acts through the circumference of a circle with a radius equal to the length of the rod. Understanding this relationship is crucial for accurately calculating the work done.
manojanand
Messages
2
Reaction score
0
The sum is a horizontal rod of mass m, len l is pivoted at one end so that it is free to rotate about a vertical axis...and a steady force fo F units acts at the free end, maintaining a direction perpendiculat to the rod. Wat is the work done by this force during the time tat the rod takeds to complete one turn, assuming that it starts from rest?
i kno its silly to ask this question but still...pls answer
 
Physics news on Phys.org
Consider the definition of work as a force acting through a distance. Over what distance does the applied force act as it completes one turn?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top