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January 30, 2015

January 30, 2015

Posted by **Susan** on Friday, December 3, 2010 at 2:00pm.

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(a) If the original length L of the rubber band is 1 m, by how much (in m) will it be stretched by the rotation?

Delta L = m

HELP: The centripetal force has to equal the elastic force.

HELP: The centripetal force is equal to

Fc = mass*length*omega2,

where length is now the stretched length, that is,

length = L + (Delta L).

The elastic force is equal to

Felast = k*(Delta L).

Since the two are equal, you have an equation for Delta L.

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(b) How much energy do you need to start the stone from rest and rotate as in part (a) with an angular speed of 9.2 rad/s? Neglect the mass of the rubber band.

Etotal = J

HELP: Add up the kinetic energy and the elastic potential energy.

HELP: Ekin = 1/2*m*v2, with v = length*omega.

Remember again to use the stretched length:

Eelast = 1/2*k*(Delta L)2

- Physics -
**Cynthia**, Sunday, December 5, 2010 at 2:01amm*(L+Delta L)*w^2=k*(Delta L)

Delta L=(m*w^2)/(k-m*w^2)

Mass: 156 g=.156kg

Answer: 0.128

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