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A 156 g hockey puck is attached to a rubber band and rotated with an angular speed of 9.2 rad/s on frictionless horizontal ice. It takes a force of 1.22 N to stretch the rubber band by 1 cm.

(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.

(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 -

    m*(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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