Easiest Cylinder to Balance - angular acceleration

AI Thread Summary
The discussion centers on determining which of three uniform cylinders is easiest to balance based on their mass, length, and moment of inertia. The correct answer is that the bamboo cylinder is easiest to balance due to its length, which provides a greater distance for the gravitational force to act, increasing net torque. Participants express confusion about the relationship between moment of inertia and angular acceleration, noting that larger moments of inertia typically result in slower angular acceleration. The conversation highlights that taller objects, like the bamboo cylinder, fall more slowly, allowing for better correction when balancing. Ultimately, the balance is influenced more by the height and torque dynamics than by the moment of inertia alone.
mike115
Messages
23
Reaction score
0

Homework Statement


24. A person tries to balance vertically each of three long uniform cylinders with equal radii on the end of his finger:


* A wood cylinder of mass m and length L with moment of inertia I
* A bamboo cylinder of mass m and length 3L with moment of inertia 9I
* A concrete cylinder of mass 4m and length 2L with moment of inertia 16I


Which object is the easiest to balance and why?

A. The wood cylinder is easiest to balance because it has the smallest moment of inertia.
B. The concrete cylinder is easiest to balance because it has the largest moment of inertia.
C. The concrete cylinder is easiest to balance because it is the most massive.
D. The bamboo cylinder is easiest to balance because it has the least density.
E. The bamboo cylinder is easiest to balance because it is longest.

Correct Answer: E

Homework Equations


net torque = I*alpha

The Attempt at a Solution


I don't understand how the answer is E. I thought the object with the largest moment of inertia would have the smallest angular acceleration. It just seems to me that if the cylinder is longer, the gravitational force will act at a further distance from the support point which would only increase the net torque.
 
Physics news on Phys.org
It is actually harder to balance small objects (such as pencils) than large objects (such as broom handles). Tall objects have a higher centre of mass. Basically this means that they will begin falling slowly, giving you a little more time to quickly correct your body to keep the object upright.
 
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 '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