Is Angular Momentum Conserved During the Changing of the String Length?

AI Thread Summary
Angular momentum is conserved when the length of the string is shortened, as the mass moves in a circular path. The initial angular momentum can be expressed as L = m * v * R, and when the string length is halved, the new angular momentum must equal the initial angular momentum. The speed of the object remains constant despite the change in string length, which implies that the angular velocity will increase to maintain conservation of angular momentum. The new period can be calculated using the adjusted radius, but the initial calculations leading to 0.5 seconds are incorrect due to misunderstanding the relationship between angular velocity and period. Understanding these principles is crucial for solving problems related to rotating systems.
AlexanderIV

Homework Statement


An object with mass m is attached to a string with initial length R, and moves on a frictionless table in a circular orbit with center C as shown in the figure. The string is also attached to the center, but its length is adjustable during the motion. The object initially moves with velocity v and angular velocity ω.
Given: m = 500 g, v = π m/s, R = 50 cm
If the length of the string is shortened from R to r = R/2 while the mass is moving, what will be the new period in SI units?

Homework Equations


T = (2πR) / v

The Attempt at a Solution


T = (2πR) / v = (2π0.5) / π = 1
=> T = (2πR/2) / v = 0.5 s

But apparently 0.5 is not the correct answer and I do not understand why.
 
Physics news on Phys.org
AlexanderIV said:
T = (2πR) / v = (2π0.5) / π = 1
=> T = (2πR/2) / v = 0.5 s

But apparently 0.5 is not the correct answer and I do not understand why.
Does the speed change when the string is shortened?
 
TSny said:
Does the speed change when the string is shortened?

No, it doesn't.
 
AlexanderIV said:
No, it doesn't.
Can you think of any physical quantity that remains conserved during the changing of the length of the string? (Maybe it's something you have recently covered in your course.)
 
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