Consider a “standard spring” : it has spring constant k, unstretched length l 0
ID: 2269225 • Letter: C
Question
Consider a “standard spring” : it has spring constant k, unstretched length l0, and cannot bend, i.e. it can only stretch or contract. (The spring could be wound around a rigid rod, for example.) One end of this spring is attached to an axle located at the origin; the other end is attached to a point mass m. A motor attached to the axle causes the spring to rotate in the xy-plane with constant angular velocity . Since the spring cannot bend, it always points radially away from the origin. The xy-plane is a frictionless horizontal table so ignore gravity.
For this problem, you must use Method A. (You’ll see why in the later parts). Reminder: you already practiced this method in Discussion 7 and Lecture 8B.
(a) Using cylindrical coordinates (s,) for the mass m, find the point s where the mass is in radial equilibrium (i.e. if it is placed at that radius with zero radial speed, it will remain at that radius).
(b) Find the condition on that will make s a point of stable equilibrium and find the frequency of small radial oscillations of the mass m around s . Also write down the general solution s(t) for the small-oscillation motion, using the symbols s and to simplify your expression.
(c) If the condition on that you found in part (b) is violated, explain why the system actually has no point of equilibrium at all (not even an unstable one). Hint: if you think there is an equilibrium point, try sketching it.
(d) Why couldn’t we use Method B? Construct T and U for the mass and try calculating the equilibrium position s using (1) the familiar equilibrium condition U(s) = 0 from PHYS 211, and (2) our new formula U eff (s) = 0 . You should find that neither of them gives the correct answer!
FYI: Even though it didn’t work, you should have obtained an effective potential that is not the same as the actual potential. This problem illustrates the typical source of Ueff U : Ueff incorporates a term from the kinetic energy that doesn’t depend on q! (only on q ). In our rotating-spring system, the forced rotation at constant angular velocity is responsible for this term.
(e) What went wrong? Identify which of the three “Method B conditions” on page 1 does not hold and explain the physical origin of the condition’s violation in this system.
PLEASE SHOW ALL WORK
Problem 1: Rotating Spring Consider a “standard spring, : t has spring constant k, unstretched length 10, and cannot bend, ie. t can only stretch or contract. (The spring could be wound around a rigid rod, for example.) One end of this spring is attached to an axle located at the origin; the other end is attached to a point mass m. A motor attached to the axle causes the spring to rotate in the xy-plane with constant angular velocity . Since the spring cannot bend it always points radially away from the origin. The xy-plane is a frictionless horizontal table so ignore gravity For this problem, you must use Method A. (You'll see why in the later parts). Reminder: you already practiced this method in Discussion 7 and Lecture 8HB (a) Using cylindrical coordinates (s,) for the mass m, find the point s where the mass is in radial equilibrium (i.e. if it is placed at that radius with zero radial speed, it will remain at that radius) (b) Find the condition on that will make a point of stable equilibrium and find the frequency of small radial oscillations of the mass m around T. Also write down the general solution s(t) for the small-oscillation motion, using the symbols s and to simplify your expression (c) If the condition on that you found in part (b) is violated, explain why the system actually has no point of equilibrium at all (not even an unstable one). Hint: if you think there is an equilibrium point, try sketching it (d) Why couldn't we use Method B? Construct T and U for the mass and try calculating the equilibrium position-using (1) the familiar equilibrium condition U,(5)= 0 from PHYS 211 , and (2) our new formula U." (5)=0. You should find that neither of them gives the correct answer! FYI: Even though it didn't work, you should have obtained an effective potential that is not the same as the actual potential. This problem illustrates the typical source of Ueff U : Ueff incorporates a term rom the kinetic energy that doesn't depend on q (only on q). In our rotating-spring system, the forced rotation at constant angular velocity is responsible for this term (e) What went wrong? Identify which of the three "Method B conditions" on page 1 does not hold and explain the physical origin of the condition's violation in this systemExplanation / Answer
given standard spring of spring constant k
unstreched length = lo
cannot bend
so while rotating with angular speed w
mass of the body = m
a. hence in cylindrical coordinates
from force balance
k(s - lo) = mw^2*s
where w = d(phi)/dt
for constant w
ks - mw^2*s = klo
s = klo/(k - mw^2)
b. now, for a stable equilibrium
dE/ds = 0
but E = 0.5*k(s - lo)^2 + 0.5*m*s^2*w^2
hence
dE/ds = k(s - lo) + m*w^2*s = 0
k(s) = mw^2*s + klo
s = klo/(k - mw^2)
as this is the result from the previous case too, hence all the equilibrium points are stabel equilibrium points
hence frequency of small osscilation = f
f = 2*pi*sqroot(k/m)
c. if consdition in b is not satisfied, then condition in A is not sartisfied meaning that there is no force balance, and hence the mass is not in equilibrium
d. if we use U = 0.5*k(s - lo)^2
dU/ds = 0
k(s - lo) = 0
s = lo ( which is already in equilibrium)
effective potential , U = 0.5k(s - lo)^2 + mw^2*r