What Speed Keeps the Cylinder Stationary in the Puck-Cylinder System?

AI Thread Summary
The discussion focuses on determining the speed required to keep a hanging cylinder stationary in a puck-cylinder system. The participant initially deduced that the net force on the hanging cylinder is zero, leading to the equation T - Fg = 0. They attempted to relate the tension and centripetal force but overlooked the need to account for both masses separately in their calculations. The correct approach involves equating the centripetal force to the gravitational force and substituting the respective formulas for each mass. The participant acknowledges the mistake and seeks clarification on how to properly incorporate both masses into the solution.
raptik
Messages
21
Reaction score
0

Homework Statement


A puck of mass m = 1.10 kg slides in a circle of radius r = 18.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 3.00 kg by a cord through a hole in the table. What speed (in m/s) keeps the cylinder at rest?

Homework Equations


F = ma

The Attempt at a Solution


I know the Fnet,y for the hanging cylinder is zero because it does not move, so I deduced that T - Fg = 0.
On the puck the 2 things I looked at was the a moving towards the hole and the tension which I am also deducing is moving towards the hole, away from the puck. (Note: I left out FN and Fg because I didn't see how they would apply in the equation for the puck.) I thus got the equation T = ma.

I then plugged in the in this T into the previous equation to get: ma - Fg = 0
I then simplified the equation to a = g. Knowing that a in circular motion with constant velocity is a = v2/R, so the equation I ended up with was v = √Rg. This came out to 1.328 ms-1, which is wrong. Could somebody tell me where my thinking went astray and how I can fix this.
 
Physics news on Phys.org
There are TWO different masses, so you need to keep two different m's in your equation or replace them with the two different numbers. Your solution will be more elegant if you write Fc = Fg first (centripetal force is provided by the gravitational force), then replace them both with their detailed formulas.
 
Thank you very much. I overlooked that there were two different masses involved in the equation and just worked them out of my problem by mistake.
 
Most welcome! It is a pleasure to save you the frustration of a tiny mistake after you demonstrate your mastery of the whole problem.
 
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