Find moment of Inertia from Force, radius and acceleration

AI Thread Summary
The discussion revolves around calculating the moment of inertia for a disc-shaped object subjected to a tangential force. The initial attempt incorrectly applied linear equations instead of rotational dynamics. The correct approach involves using Newton's second law for rotation, expressed as torque (r * F) equating to the product of moment of inertia (I) and angular acceleration (α). The final formula derived is I = rF/α, clarifying the relationship between force, radius, and angular acceleration. Understanding the distinction between linear and angular quantities is crucial for solving such problems.
iva
Messages
21
Reaction score
1

Homework Statement




A disc shaped object is made of a non-uniform material. Its radius is r and it is fre to rotate about an axis through its centre. If a force F applied tangentially at the edge of the object produces the angular acceleration a, what is its moment of inertia for rotation about the axis?


Homework Equations



I=mr2
F=ma

The Attempt at a Solution



If F=ma then m=F/a

so that i=mr2 becomes I=Fr2/a

The answer in the book is I=Fr/a

Where did i go wrong? is it something to do with the force being tangent to the disc that i didn't do something?

Thanks!
 
Physics news on Phys.org
iva said:
If F=ma then m=F/a
Careful. "a" is the angular acceleration, not the linear acceleration. (It's better to use alpha for angular acceleration.) What's Newton's law for rotation?

so that i=mr2 becomes I=Fr2/a
That formula for moment of inertia does not apply here.

All you need is Newton's 2nd law for rotation.
 
Thanks i get it now,

So all i really needed was 1 equation:

rotational force equation: r* F=I * alpha so I=rF/alpha right?
 
Right!
 

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

Moment of Inertia with pulley and two masses on a string (iWTSE.org)
Replies
8
Views
12K
Solving for 'a' with Torque, Force, and Mass Moment of Inertia
Replies
3
Views
2K
Calculating Moment of Inertia for Hollow Cylinder w/ Water
Replies
40
Views
5K
Moment of inertia of a cuboid parallel to one of its faces (repost with diagram)
Replies
25
Views
2K
Torque about an accelerating point
Replies
5
Views
3K
Parallel Axis Theorem Not Working
Replies
28
Views
1K
Calculating Moment of Inertia from Ball Accelerations
Replies
1
Views
1K
Moment of inertia of T bar about 3 axes
Replies
21
Views
2K
Moment of inertia of a disc using rods as differential elements
Replies
8
Views
2K
Angular acceleration of plank hanging off edge with a weight
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top