How Do You Calculate the Time for a Mass to Fall Using Moment of Inertia?

  • Thread starter Thread starter jago-k1
  • Start date Start date
  • Tags Tags
    Moment
AI Thread Summary
The discussion focuses on calculating the time it takes for a hanging mass to fall 31.5 cm using the moment of inertia. The user presents their attempts at solving related problems, including calculating the moment of inertia for a two-mass system and determining rotational kinetic energy. They correctly compute the moment of inertia for a support rod and shaft and seek assistance in deriving the time for the mass to fall. The thread emphasizes the application of physics equations related to energy and motion, highlighting the user's need for clarification on the final calculation.
jago-k1
Messages
13
Reaction score
0

Homework Statement


How long does it take for the hanging mass to fall the same distance 31.5 cm?
I don't know how to solve for T with everything I have. Help please.

Homework Equations



mgh=(1/2)I(2h/tr)2 + (1/2)m(2h/t)2

The Attempt at a Solution



Okay, previous questions are[All correct]
1.
Two masses of 150. g are suspended from a massless rod at a distance of 9.0 cm from the center. What is the moment of inertial of the two-mass system about the center of the rod?

I=m1*R^2
since they are two masses that weigh the same
I= 2(.15kg)(.09m)^2=2.43E-03 Correct


2.
If the masses rotate with an angular velocity of 2.45 rad/s, what is the rotational kinetic energy of the system?

KEr=(1/2)Iw^2 = (2.43E-03 )(2.45 rad/s)^2 = 7.29E-03 J Correct


3.
Consider the setup shown in the lab manual but with the large masses removed from the support rod. If the hanging mass is 100. g and drops a distance 31.5 cm in a time of 6.1 s, what is the moment of inertia of the support rod and shaft? The radius of the shaft is 0.50 cm.

I=mr^2(gt^2/2h - 1) = 1.446E-03 kg m^2 Correct


4.
Now two masses each of 200 g are placed on the rod at a distance of 11.0 cm from the point of rotation. What is the TOTAL moment of inertia of the masses plus rod and shaft?
Yes, Computer gets: 6.286E-03 kg m^2

I=2MR^2+Io is what you get from #3
I=2(.2kg)(.11m)^2 + Io = 6.286E-03 kg m^2 Correct
 
Physics news on Phys.org
BUMP if anyone can help out.
 
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