Moment of Inertia of Cabinet Door - 49.1x76.5cm, 550kg/m3

AI Thread Summary
The discussion revolves around calculating the moment of inertia of a cabinet door made from plywood, considering its dimensions and mass. The door measures 49.1 cm by 76.5 cm and has a thickness of 1.81 cm, with a density of 550 kg/m3. A 181-g handle is positioned 45 cm from the lower hinge, and the contribution of hinge components is ignored. Participants express confusion about the correct formula and the relationship between the door's dimensions, mass, and the moment of inertia. The conversation emphasizes the need to understand the underlying physics rather than just the dimensions.
blackfoal
Messages
4
Reaction score
0

Homework Statement


A sheet of plywood 1.81 cm thick is used to make a cabinet door 49.1 cm wide by 76.5 cm tall, with hinges mounted on the vertical edge. A small 181-g handle is mounted 45 cm from the lower hinge at the same height as that hinge. If the density of the plywood is 550 kg/m3, what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.


Homework Equations





The Attempt at a Solution


[((weight)*((tall)^2))+((density*tall*thick)*((wide^3)/3))]

answer is not true according to this formula. help me.
 
Physics news on Phys.org
You can look at this problem from a bird's eye view, since the vertical axis is of no concern to us. How is the moment of inertia of the door from the hinges to handle related to the distance from the hinge to the handle and the door's mass?

I think you're getting confused here with the door's dimensions. Think less of this and more of the physics behind it. Once you get the physics, you'll understand why they give you the length, height and thickness of the door.
 
i didnt understand anything your comment :) do you give a formula? maybe i can understand on the formula.
 
i looked but i can not solute.
 
is there anybody for helping to me.
 
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...
Back
Top