f.6 phy. circular motion

2008-10-12 11:03 pm

2) An elastic cord 1 m long when unstretched obeys Hooke's law to an extension of 0.2m,which it breaks. A rubber bung is tied to one end of the cord and then whirled around in a horizontal circle of radius 1.1m with angular velocity 6rad/s. The centripetal force required is 5 N. Assume the cord is horizontal.

Find force constant of the elastic cord and total energy of rotating system.

If the rubber bung is then acc. at 0.2 rad in what time will the elastic cord break?

回答 (1)

Audrey Hepburn

2008-10-12 11:38 pm

✔ 最佳答案

2. By Hooke's law

kx = 5

k(1-1 - 1) = 5

Force constant, k = 50 Nm-1

Linear speed, v = radius X angular velocity = 1.1 X 6 = 6.6 ms-1

By 5 = m(1.1)(6)2

Mass of the bung, m = 0.126 kg

Total energy in the rotating system

= E.P.E. stored in the cord + K.E. of the bung

= 1/2 kx2 + 1/2 mv2

= 1/2 (50)(1.1 - 1)2 + 1/2 (0.126)(6.6)2

= 3 J

For the cord to break, required force, F = (50)(1.2 - 1) = 10 N

which accounts for the centripetal force

Hence, F = mrw2

10 = (0.126)(1 + 0.2)w2

Angular velocity when the cord breaks, w = 8.124 rads-1

By w = ω + αt

8.124 = 6 + 0.2t

Time, t = 10.6 s

參考: Myself~~~

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