Newton's Second Law & Moment of Inertia Derivation?

AI Thread Summary
The discussion focuses on deriving the expression for the moment of inertia using Newton's Second Law, referencing a lab experiment. Participants clarify the relationship between force, mass, and acceleration, as well as the formula for torque. There is a specific attempt to manipulate equations involving effective mass and gravitational acceleration to reach the desired expression. The necessity of calculus for understanding the problem is debated, with some suggesting it may not be essential. The physical system in question is confirmed to be identical to that used in the lab experiment.
danyalasdf
Messages
6
Reaction score
0

Homework Statement


We did something very similar to this in lab

http://webenhanced.lbcc.edu/physte/phys2ate/2A LAB HANDOUTS/Moment of Inertia.pdf

Use Newton's Second to derive the expression for the experimentally determined moment of inertia

Homework Equations


Newton's Second Law is F=m*a

s=(1/2)*a*t^2

Experimentally Moment of Inertia

I=r^2(m((gt^2/2s)-t) - mf)

Trying to get to this ^

mf= mass effective not much meaning just mass in kg
If it confusing the gt^2 is divided by 2s then it is subtracted by t and multiplied by r^2 and then minus mf

Torque= F*r= m*r*a

T= (mf + m)(g - a) = tension

The Attempt at a Solution



T= (mf + m)(g - (2*s/t^2))

T= (mf + m)((1/2)(a*t^2) - a)

T= (mf + m)(a((1/2)*t^2 - 1)

I am stuck write here
 
Last edited by a moderator:
Physics news on Phys.org
Is calculus is a prerequisite for this course?
 
Pythagorean said:
Is calculus is a prerequisite for this course?

yes it is Physics Calculus Based 2211K
 
Calc may not be necessary.

So what's the physical system you're doing in this problem? Is it the same exact system as in the lab?
 
Pythagorean said:
Calc may not be necessary.

So what's the physical system you're doing in this problem? Is it the same exact system as in the lab?

yes it is the exact same system
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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