Problem 70 - University Physics with Modern Physics

Question

CP A child with poor table manners is sliding his \(250-\mathrm{g}\) dinner plate back and forth in \(\mathrm{SHM}\) with an amplitude of 0.100 \(\mathrm{m}\) on a horizontal surface. At a point 0.060 \(\mathrm{m}\) away from equilibrium, the speed of the plate is 0.400 \(\mathrm{m} / \mathrm{s}\) . (a) What is the period? (b) What is the displacement when the speed is 0.160 \(\mathrm{m} / \mathrm{s} ?\) (c) In the center of the dinner plate is a 10.0 -g carrot slice. If the carrot slice is just on the verge of slipping at the endpoint of the path, what is the coefficient of static friction between the carrot slice and the plate?

Step-by-Step Solution

Answer

(a) 1.0 s; (b) 0.0947 m; (c) 0.0638.

Step 1: Use Conservation of Energy

First, apply the principle of conservation of mechanical energy for simple harmonic motion. The total energy in SHM is given by the equation:\[ E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \]where \(m = 0.25\, \text{kg}\), \(v = 0.400\, \text{m/s}\), \(x = 0.060\, \text{m}\), \(A = 0.100\, \text{m}\), and \(k\) is the spring constant. Substitute the known values to find \(k\). Solving will yield:\[ 0.5 (0.25)(0.4)^2 + 0.5 k (0.06)^2 = 0.5 k (0.1)^2 \]\[ 0.02 + 0.0018 k = 0.005k \]Solve for \(k\):\[ k = \frac{0.02}{0.0032} = 6.25\, \text{N/m} \]

Step 2: Calculate the Period of Motion

The period \(T\) of SHM is given by:\[ T = 2\pi \sqrt{\frac{m}{k}} \]Substitute the values of \(m = 0.25\, \text{kg}\) and \(k = 6.25\, \text{N/m}\):\[ T = 2\pi \sqrt{\frac{0.25}{6.25}} \approx 1.0\, \text{s} \]

Step 3: Find the Displacement at a Given Speed

When the speed is \(v = 0.160\, \text{m/s}\), use energy conservation again:\[ \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \]Substitute for \(v = 0.160\, \text{m/s}\):\[ 0.5 (0.25) (0.16)^2 + 0.5 (6.25) x^2 = 0.5 (6.25) (0.1)^2 \]\[ 0.0032 + 3.125x^2 = 0.03125 \]Solve for \(x\):\[ 3.125x^2 = 0.03125 - 0.0032 \]\[ 3.125x^2 = 0.02805 \]\[ x^2 = \frac{0.02805}{3.125} \approx 0.008976 \]\[ x \approx 0.0947\, \text{m} \]

Step 4: Find Coefficient of Static Friction

At the endpoint of the path, the carrot slice is on the verge of slipping, which means the maximum static friction equals \(mg\) times the static friction coefficient. The maximum acceleration is at this endpoint:\[ a = \omega^2 A = \frac{k}{m} A \]\(a = 6.25 \times 0.1 = 0.625\, \text{m/s}^2\)Using \(f_s = ma\) gives:\(\mu_s mg = ma\)\(\mu_s = \frac{a}{g} = \frac{0.625}{9.8} \approx 0.0638\)

Key Concepts

Conservation of Energy

In Simple Harmonic Motion (SHM), the conservation of energy plays a pivotal role. This principle states that the total mechanical energy in a system remains constant as the object oscillates. We can split this energy into two parts:

• Kinetic Energy (\(KE = \frac{1}{2}mv^2\)) which is energy due to motion.
• Potential Energy (\(PE = \frac{1}{2}kx^2\)) which is energy stored in the system due to position.

For the system of the dinner plate sliding back and forth, this principle helps to understand how energy shifts between kinetic and potential forms. When the plate is at the equilibrium position, the energy is completely kinetic. At maximum displacement, the energy is entirely potential. By equating the total energy at any two points, we get:\[KE + PE = \frac{1}{2}kA^2\]This equation is crucial in problems where speed or position at a given time is unknown. By substituting known values, one can solve for the unknowns such as the spring constant or displacement.

Static Friction

Static friction is the force that keeps an object at rest when it is subjected to an opposing force. When dealing with SHM, static friction becomes a key topic, especially when an object is on the brink of slipping, like the carrot slice on the dinner plate.
The maximum static friction is described by:\(f_s = \mu_s N\) where \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force. In this scenario, if the carrot is about to slide, the force of static friction is exactly balanced by the maximum outward force due to SHM.
To find the coefficient of static friction, remember that maximum frictional force equals the product of mass (\(m\)), the maximum acceleration (\(a\)), and the gravitational force (\(g\)). Therefore:\(\mu_s = \frac{a}{g}\)Applying this expression yields the coefficient, illustrating the delicate balance maintained by static friction.

Period of Motion

The period of motion is a fundamental concept in SHM, representing the time it takes for one complete cycle of motion. Denoted by \(T\), it remains constant for a given system and can be determined by the properties of the mass and the spring.
For a system obeying Hooke's law, the formula for period is:\[T = 2\pi \sqrt{\frac{m}{k}}\]where \(m\) is the mass, and \(k\) is the spring constant. The period is independent of amplitude and gives us valuable insights into how quickly or slowly the system oscillates.
By using this formula, the period can easily be calculated once the mass and spring constant are known, as demonstrated in the solution to the exercise.

Spring Constant

The spring constant (\(k\)) is a measure of the stiffness of the spring in an SHM system. The larger the spring constant, the harder it is to compress or stretch the spring. It plays a crucial role in determining the oscillatory behavior of a system.
In our problem, the spring constant appears in several formulas, notably in connecting the energy, the period of motion, and the potential forces in the system. By using the conservation of energy principle, \(k\) is calculated based on kinetic energy at certain points and the known amplitude.
Understanding \(k\) helps to predict how responsive a system will be to external forces. Essentially, knowing \(k\) allows us to see how tightly or loosely linked the plate's motion is to its elastic medium, unlocking the dynamics of motion.