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
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.
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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
1 answer
1 answer