Asked by Help
Rubber Band
A 156 g hockey puck is attached to a rubber band and rotated with an angular speed of 9.8 rad/s on frictionless horizontal ice. It takes a force of 1.12 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
.154 OK
(b) How much energy do you need to start the stone from rest and rotate as in part (a) with an angular speed of 9.8 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
A 156 g hockey puck is attached to a rubber band and rotated with an angular speed of 9.8 rad/s on frictionless horizontal ice. It takes a force of 1.12 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
.154 OK
(b) How much energy do you need to start the stone from rest and rotate as in part (a) with an angular speed of 9.8 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
Answers
Answered by
bobpursley
centripetal force= masspuck*w^2*r
a) stretch= centripetal force/1.12cm
b) total energy= 1/2 mass*w^2*r^2+1/2 (1.12/cm)*stretch^2
a) stretch= centripetal force/1.12cm
b) total energy= 1/2 mass*w^2*r^2+1/2 (1.12/cm)*stretch^2
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