Speed of Cylinder & Hoop Down Ramp: Analysis

AI Thread Summary
The discussion focuses on calculating the speed of a cylinder and a hoop rolling down a ramp using conservation of energy principles. It emphasizes the importance of including rotational kinetic energy in the equations if the objects are allowed to roll. The potential energy at the top of the ramp is equated to the sum of translational and rotational kinetic energies at the bottom. The derived formula for speed incorporates the moment of inertia and the radius of the objects. This approach allows for specific cases to be analyzed for both a solid cylinder and a hoop.
missnuss
Messages
8
Reaction score
0

Homework Statement



Consider both a cylinder of radius Rc and mass mc and a hoop of radius Rh and mass mh. If both are at the top of a ramp of length L and at an angle theta what are the objects speed at the bottom of the ramp?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
You have to show your work before people help you. That IS the modus operandi of this
forum. Check thread started by me for example...
 
Last edited:
Sorry. I didn't post what I had tried to do because I have no clue where to even start. I'm not looking for the answer, I just wanted some help.
 
Use conservation of energy. Are the cylinders "allowed" to roll down the ramp? If so, you have to add in a rotational kinetic energy term into your equation.
 
If L is the length of the ramp, then the bottom point of any symmetric object would be
at height L\sin \theta. Using conservation of energy, the potential energy
at the top should be equal to the sum of translational and rotational kinetic energies
at the bottom. For an object with moment of inertia about its center of mass I,

mg(L \sin \theta)= \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2

for pure rolling motion, \omega = v/R so

mg(L \sin \theta)= \frac{1}{2}mv^2 + \frac{1}{2}I \frac{v^2}{R^2}

solving for v, we have

v = \left[\frac{2gL \sin \theta}{1+(I/mR^2)}\right]^{1/2}

from here, you can derive special cases for different bodies, in your case, a solid
cylinder (or is it hollow) and a hoop.
 
Thank you so very much
 
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