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
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) To calculate the stretch in the rubber band due to rotation, we can use Hooke's Law, which states that the force required to stretch or compress a spring (or in this case, a rubber band) is directly proportional to the displacement or change in length.
In this case, the force required to stretch the rubber band by 1 cm is given as 1.12 N. Let's convert the displacement to meters: 1 cm = 0.01 m.
Using Hooke's Law, we can write the equation:
Force = k * displacement
Where:
- Force is the force required to stretch the rubber band (1.12 N)
- k is the spring constant of the rubber band (which we need to calculate)
- displacement is the change in length of the rubber band (which we want to find)
Simplifying the equation, we have:
displacement = Force / k
Now, we can rearrange the equation to solve for k:
k = Force / displacement
Substituting the given values, we have:
k = 1.12 N / 0.01 m
Evaluating this expression, we find:
k ≈ 112 N/m
Now, let's calculate the stretch in the rubber band when the hockey puck rotates.
Using the formula for the change in length of the rubber band due to rotation:
ΔL = m * ΔL / k
Where:
- ΔL is the change in length of the rubber band
- m is the mass of the hockey puck attached to the rubber band (156 g = 0.156 kg)
- ΔL is the displacement, which we want to find
- k is the spring constant of the rubber band (112 N/m)
Substituting the values, we have:
ΔL = (0.156 kg * 9.8 m/s^2) / (112 N/m)
Evaluating this expression, we find:
ΔL ≈ 0.0137 m
Therefore, the rubber band will be stretched by approximately 0.0137 meters due to rotation.
(b) To calculate the energy required to start the stone from rest and rotate as in part (a), we need to add up the kinetic energy and the elastic potential energy.
The kinetic energy can be calculated using the formula:
Ekin = 1/2 * m * v^2
Where:
- Ekin is the kinetic energy
- m is the mass of the hockey puck (156 g = 0.156 kg)
- v is the velocity of the hockey puck, which can be calculated from the angular speed (ω) and the stretched length (ΔL) using the formula v = ΔL * ω
Substituting the values, we have:
v = 0.0137 m * 9.8 rad/s
Evaluating this expression, we find:
v ≈ 0.134 m/s
Now we can calculate the kinetic energy:
Ekin = 1/2 * 0.156 kg * (0.134 m/s)^2
Evaluating this expression, we find:
Ekin ≈ 0.00112 J
The elastic potential energy can be calculated using the formula:
Eelast = 1/2 * k * (ΔL)^2
Where:
- Eelast is the elastic potential energy
- k is the spring constant of the rubber band (112 N/m)
- ΔL is the change in length of the rubber band (0.0137 m)
Substituting the values, we have:
Eelast = 1/2 * 112 N/m * (0.0137 m)^2
Evaluating this expression, we find:
Eelast ≈ 0.0106 J
Finally, we can calculate the total energy:
Etotal = Ekin + Eelast
Substituting the values, we have:
Etotal ≈ 0.00112 J + 0.0106 J
Evaluating this expression, we find:
Etotal ≈ 0.0117 J
Therefore, the total energy required to start the stone from rest and rotate as in part (a) with an angular speed of 9.8 rad/s is approximately 0.0117 Joules.