A spring, with R1 = 8.88 cm, has a spring constant of k = 3.14Newtons per centim

Question

A spring, with R1 = 8.88 cm, has a spring constant of k = 3.14Newtons per centimeter (N/cm). A mass, m, is attached to the springand rotates about the point where the spring is anchroed. As aresult the spring stretches to 12.00 cm. (a) What is the magnitudeof the centripetal force exerted by the spring? (b) If thefrequency of rotation in part (a) above is 300 RPM, what is thecentripetal acceleration? I got this question wrong on my homework. My teacher did nottell us the answer nor showed us how to do it. I want to understandwhat i did wrong so I won't make the same mistake in the futureexams. Can someone please do this out for me. Thanks. A spring, with R1 = 8.88 cm, has a spring constant of k = 3.14Newtons per centimeter (N/cm). A mass, m, is attached to the springand rotates about the point where the spring is anchroed. As aresult the spring stretches to 12.00 cm. (a) What is the magnitudeof the centripetal force exerted by the spring? (b) If thefrequency of rotation in part (a) above is 300 RPM, what is thecentripetal acceleration? I got this question wrong on my homework. My teacher did nottell us the answer nor showed us how to do it. I want to understandwhat i did wrong so I won't make the same mistake in the futureexams. Can someone please do this out for me. Thanks.

Explanation / Answer

(a) if the spring stretches to 12 cm, then   x = 12 - 8.88   = 3.12cm . and the force exerted by the spring on the mass, which is thecentripetal force, is .         3.14 N/cm * 3.12 cm =    9.80Newtons . (b) now,    rotation = 300rpm = 5 revolutions per second. . This means the period of rotation is   T = 0.2seconds . And you know    speed = distance /time     so    v = 2 r /T = 2 * 0.12 / 0.2 =   3.770m/s . And        centrip acc=   v2 / r    =   3.7702 / 0.12 =    118.44m/s2

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