What is the Angular Speed of a Rotating Hoop with Given Radius and Mass?

AI Thread Summary
The discussion revolves around calculating the angular speed of a rotating hoop with a radius of 8.00 cm and mass of 0.180 kg after it descends 95.0 cm. The initial approach incorrectly assumed that kinetic energy was zero, leading to confusion about angular speed. The user attempted to apply formulas related to a yoyo, but struggled to connect angular speed and linear speed due to a lack of known variables. The descent of the hoop is influenced by gravitational force and its moment of inertia, complicating the calculations. Ultimately, the user seeks clarification on deriving the angular speed and the speed of the hoop's center.
Heat
Messages
272
Reaction score
0

Homework Statement



A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180kg. The free end of the string is held in place and the hoop is released from rest (the figure View Figure ). After the hoop has descended 95.0cm, calculate

the angular speed of the rotating hoop and the speed of its center.


Homework Equations





The Attempt at a Solution



First attempt was like this

I thought that KE is 0 at initial ,and thought that I could get angular speed this way, but that would just give me that angular speed is 0. :O

Then I tried a formula given by the book for a yoyo, which is vcm = sqrt (4/3gh)
that did not result, and then is another formula that vcm = R(omega)

I don't have omega, and don't have vcm. So I can't use it at this instance. :(
 
Physics news on Phys.org
The hoop is decending under the force of gravity by virtue of its weight, mg. At the same time, it is unraveling the string. It's vertical velocity is the same as the tangential velocity (speed) at it's radius. And then there is the rotational inertia to consider.

For moments of inertia and rotational motion equations, see -

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
 
i don't get it..so the angular speed of the rotating hoop w = ?

and

the angular speed of the speed of its center v = Square_Root_Of_((4/3)gh) ... ??
 
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