A particle of mass m moves under an attractive central force F(r) = ? Kr, with a

Question

A particle of mass m moves under an attractive central force F(r) = ? Kr, with angular momentum L. For circular motion, determine the energy and the radius of the circle.

(A) Now calculate the frequency of small radial oscillations away from the circular orbit.
Data: m = 0.36 kg; K = 14.00 N/m; L = 23.20 kg m^2/s

F = -Kr so U(r) = -kr^2/2
dUeff/dr = -L^2/(mr^3)+kr (1st derivative of Ueff) (1)
dUeff^2/d^2r = 3L^2/mr^4+k (second derivative of Ueff) (2)

to be stable dUeff/dr = 0 so r^4 = L^2/(mk) (3)

plugging (3) into (2)

dUeff^2/d^2r = 4k

the formula for frequency is w(r) = sqrt(dUeff^2/d^2r/m) (I think)

w(r) = sqrt(4k/m) = 12.47 Hz but this is wrong! No idea where I went wrong.

Explanation / Answer

L^2/m*r^3=Kr L^2/m*K=r^4 r=3.2m now u proceed the same

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.