Rotational inertia of a rod that has mass on one end

AI Thread Summary
The discussion focuses on calculating the rotational inertia of a system consisting of a metal rod and a mass attached to one end. The rod has a mass of 2 kg and a length of 4 meters, while the mass at the end is 3 kg. The rotational inertia of the rod alone is calculated to be 10.6 kg·m², but the effect of the 3 kg mass on the total inertia is unclear. It is noted that the total rotational inertia of the system can be found by summing the inertia of the rod and the inertia of the mass treated as a point particle at the end of the rod. Understanding how to combine these components is crucial for solving the problem accurately.
xd14
Messages
1
Reaction score
0

Homework Statement


a 3 kg mass is on the end of a metal rod which is pivoted at one end. the mass of the rod is 2kg its length is 4 meters

Homework Equations


I=(ML^2)/3

The Attempt at a Solution


the rotational inertia of the rod itself is 10.6 but i don't know how the 3kg mass at the end would effect things
 
Physics news on Phys.org
xd14 said:

Homework Statement


a 3 kg mass is on the end of a metal rod which is pivoted at one end. the mass of the rod is 2kg its length is 4 meters

Homework Equations


I=(ML^2)/3

The Attempt at a Solution


the rotational inertia of the rod itself is 10.6 but i don't know how the 3kg mass at the end would effect things
Assuming the mass at the end has negligible size , what is the I of a particle of mass m about a pivot point , and then note the I of the rod-mass system is the sum of the I's of its parts.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Correct Use of the Parallel Axis Theorem for Moment of Inertia
Replies
2
Views
2K
Uniform rotating rod about a point at one end of the rod
Replies
3
Views
2K
Moment of Inertia in Planet System
Replies
9
Views
2K
Rotation of Rigid Bodies: Rotating stick with disc on top
Replies
9
Views
2K
Question about springs and rotational/translational energy
Replies
4
Views
1K
Find velocities when a mass leaves a system of a rod with two masses
Replies
10
Views
2K
Rotational inertia - globes connected by a thin rod
Replies
9
Views
2K
Oscillatory motion problem: Bicycle wheel rotation oscillation
Replies
17
Views
895
Moment of inertia of a rod bent into a square
Replies
12
Views
2K
A rod of length L and mass M has a nonuniform mass distribut
Replies
1
Views
4K

Hot Threads

Recent Insights

Back
Top