In a machine a cylinder with a mass of 80.0 g is pushed down into a hole with a spring at the
bottom as shown in Figure 1. When the spring is uncompressed the cylinder rests on top of it with
its end just at the entrance to the hole. The cylinder is pushed by a rod which is angled at 30◦ as
shown. The rod exerts a force of 10.0 N parallel to the line that the rod points along. The walls of
the hole are well oiled so you can ignore friction. The spring has a stiffness of 1200 N/m. Getting
the timing of the machine right requires that we know the speed of the cylinder when the spring is
6.00 cm long as shown.
(a) Choose your system and define your axes clearly.
(b) Draw an energy bar chart for the situation.
(c) Solve for the speed of the cylinder when the spring is 6.00 cm long.
(d) Calculate the height of each of the “bars” in your energy bar chart. Is there one of the types
of energy that you used which is very small compared to the others? Recalculate the answer
to c) neglecting this contribution to see how good the approximation is. (In practice when
objects in machines are moving fast and there are spring forces involved the approximation
you have just found is a very good approximation.)
(a) To solve this problem, we need to choose a system and define our axes. Let's choose the cylinder as our system. We can define our axes with the x-axis parallel to the rod and the y-axis perpendicular to the ground.
(b) An energy bar chart shows the initial and final energies of a system. In this case, we can identify three types of energy: gravitational potential energy (PE), kinetic energy (KE), and elastic potential energy (PEelastic). The energy bar chart can be represented as follows:
Initial Energy:
- Gravitational Potential Energy: 0 J
- Kinetic Energy: 0 J
- Elastic Potential Energy: 0 J
Final Energy:
- Gravitational Potential Energy: mgh
- Kinetic Energy: (1/2)mv^2
- Elastic Potential Energy: (1/2)kx^2
where:
m is the mass of the cylinder
g is the acceleration due to gravity
h is the height of the cylinder above the spring when it is at rest
v is the speed of the cylinder when the spring is 6.00 cm long
k is the stiffness of the spring
x is the displacement of the spring from its equilibrium position
(c) To solve for the speed of the cylinder when the spring is 6.00 cm long, we can use the principle of conservation of mechanical energy. The total energy in the system is conserved, so we equate the initial energy to the final energy:
mgh + 0 + 0 = 0 + (1/2)mv^2 + (1/2)kx^2
Since the cylinder is initially at rest, the initial kinetic energy is zero. We also know that the height of the cylinder above the spring when it is at rest is equal to the uncompressed length of the spring, which is the equilibrium position of the spring. Let's call this equilibrium position x_eq.
mgh + 0 + 0 = 0 + (1/2)mv^2 + (1/2)k(x - x_eq)^2
We can rearrange this equation to solve for v:
v^2 = 2gh + k(x - x_eq)^2
v = sqrt(2gh + k(x - x_eq)^2)
Substituting the given values, we can calculate the speed of the cylinder when the spring is 6.00 cm long.
(d) To calculate the height of each of the "bars" in the energy bar chart, we need to know the values of m, g, h, k, x, and x_eq. Unfortunately, these values are not provided in the given information. You will need to find or determine these values before calculating the energies.
If one of the types of energy is very small compared to the others, we can neglect it in our calculation as an approximation. To see how good this approximation is, recalculate the answer to part (c) neglecting the small contribution and compare it to the previous answer.