- #1
Perpendicular component of momemtum?
- Thread starter ohheytai
- Start date
-
- Tags
- Component Perpendicular
AI Thread Summary
The discussion revolves around the challenges of calculating the perpendicular component of momentum in a physics problem. The initial approach using the formula mv²/r with a fixed radius is deemed incorrect due to the variable length of the string. The radius of curvature is critical, as it relies on the elasticity of the string, which is not provided until a later part of the question. Attempts to use the relaxed length of 3.63 meters also do not yield the correct answer. The consensus is that without additional information on elasticity, a definitive solution cannot be reached.
- #2
tiny-tim
Science Advisor
Homework Helper
- 25,837
- 258
hi ohheytai! 
this is a really unpleasant question
i see that you've used mv2/r, with r = 3.7 (the length of the string), to get 214.05
that would be correct if the string was of fixed length, but unfortunately it isn't, so the radius of curvature won't be the same as the length of the string
on the basis of the information at the top of the question, i don't think there's any answer, since the answer depends on the radius of curvature, which depends on the elasticity, which you're given no clue about until part (e) (the relaxed length = 3.63m)
this is a really unpleasant question
i see that you've used mv2/r, with r = 3.7 (the length of the string), to get 214.05
that would be correct if the string was of fixed length, but unfortunately it isn't, so the radius of curvature won't be the same as the length of the string
on the basis of the information at the top of the question, i don't think there's any answer, since the answer depends on the radius of curvature, which depends on the elasticity, which you're given no clue about until part (e) (the relaxed length = 3.63m)
have you tried it using that 3.63?
- #3
ohheytai
- 81
- 0
doesnt work :(
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...
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...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
Similar threads
Hot Threads
-
Collision of a bullet on a rod-string system: query
- Started by palaphys
- Replies: 70
- Introductory Physics Homework Help
Recent Insights
-
Insights Quantum Entanglement is a Kinematic Fact, not a Dynamical Effect
- Started by Greg Bernhardt
- Replies: 7
- Quantum Physics
-
Insights What Exactly is Dirac’s Delta Function? - Insight
- Started by Greg Bernhardt
- Replies: 0
- General Math
-
Insights Relativator (Circular Slide-Rule): Simulated with Desmos - Insight
- Started by Greg Bernhardt
- Replies: 1
- Special and General Relativity
-
Insights Fixing Things Which Can Go Wrong With Complex Numbers
- Started by PAllen
- Replies: 7
- General Math
-
Insights Fermat's Last Theorem
- Started by fresh_42
- Replies: 91
- General Math
-
Insights Why Vector Spaces Explain The World: A Historical Perspective
- Started by fresh_42
- Replies: 0
- General Math