Circular, Rotational, and Harmonic Motion - AP Physics 1

Since the spring is Hookean, the relationship between the force and displacement from equilibrium of the mass can be expressed by Hooke's Law:

Since this equation is linear, the force and displacement are directly proportional. Thus, when the displacement doubles, the force doubles.

Doubling the initial distance the weight is displaced from equilibrium doubles the amplitude, but does not change the period or the phase of the oscillator. So in the general equation of an oscillator:

, only the term changes, and it doubles.

In an isolated system, there is no net torque. If there is no net torque on the system, then the total angular momentum of the system remains the same. The angular momentum of a rotating object is equal to the moment of inertia of the object multiplied by the object's angular velocity.

is the symbol for angular momentum, is the moment of inertia, and is the angular velocity.

Therefore, if the moment of inertia, , is halved, then for the angular momentum, , to remain constant, the angular velocity, , must be doubled. This is because , which is the multiplicative identity. Anything multiplied by one remains the same. So, the final angular velocity would be twice as large as it was originally.

We can begin with the expression for conservation of energy to solve this problem:

Since we are told that the system is initially at rest, we can eliminate initial kinetic energy to get expression (1):

Expanding each term from left to right to:

We are told that mass A is held at an angle of 15 degrees below the horizontal, so we can use the sine function to determine the height of each mass. We will assume that the lowest point of rotation has a height of 0.

Where d is the vertical distance below horizontal, and r is the radius from the center of the rod, which is a distance equal to half the length of the rod:

Rearranging for d:

We know that this is the distance below horizontal that mass A begins at and the distance above the horizontal that mass B begins at. Also, horizontal is at a height of half a length rod above our reference height, so we can say:

Plugging these into our expanded expression for initial potential energy, we get:

Moving on to final potential energy:

We are asked to find the maximum angular momentum achieved by mass B. This occurs when mass B is traveling at it's highest velocity, which is when the larger of 2 masses, mass A, is traveling through the low point of rotation. This means that mass A is at a height of 0, and thus has no final potential energy. We can then say:

Also, we know that mass B will be at the high point of rotation, which means that its height will be the length of the rod. Therefore, we can say:

Moving on to final kinetic energy:

We don't have to expand the velocity term since the two masses are always traveling at the same speed.

Now, substituting our expanded terms back into expression (1), we get:

In order to determine angular momentum, we are going to need to solve this equation for final velocity, so let's begin rearranging:

We have all of these values, so time to plug and chug:

This is the maximum velocity that mass B will experience. We can now use the expression for angular momentum:

where:

Where r is half the length of the rod:

We have all of these values, so time to plug and chug:

To calculate torque, this equation is needed:

Next, identify the given information:

Plug these numbers into the equation to determine the torque:

We can determine the vertical components by using some geometry.

Since this problem boils down to being a problem of conservation of energy, we can state that

Since the mass is released from rest and we can state at the bottom of its arc it will be at height 0 m, this equation can be simplified:

is equal to the height of the mass above 0 m; therefore, by referring to the diagram, we can determine that . By plugging this in and rearranging the equation, we can solve for the speed of the mass when the string is vertically aligned:

When the object is released, it has no kinetic energy (it isn't moving) and a potential energy of

When the object passes through the equilibrium point, it has no potential energy () and a kinetic energy of

Due to the conservation of energy, these two quantities must be equal to each other:

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

The force applied by the spring in the opposite direction must be equal to this:

The angular velocity of the harmonic system is equal to the square root of the spring constant over the mass:

Since we need the angular velocity to be and we are given the spring constant as , then we can set up an equation:

The velocity equation can be found by differentiating the position equation.

Here, we made use of the chain rule in taking the derivative.

This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:

According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:

Subsituting in the expressions for potential and kinetic energy, we get:

We can eliminate mass to get:

Rearranging for final velocity, we get:

In order to solve for the velocity, we need to find the initial height of the ball.

The following diagram will help visualize the system:

From this, we can write:

Using the length of string and the angle it's held at, we can solve for :

Now that we have all of our information, we can solve for the final velocity:

Doubling the length of a pendulum increases the period, so it decreases the frequency of the pendulum. The frequency depends upon the square root of the length, so the frequency decreases by a factor of . Neither of the other parameters (amplitude, phase) change.

We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:

In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:

Dividng one expression by the other, we get a ratio:

We know that , so we can rewrite the expression as:

Rearranging for P2, we get:

We need to know the formula for the period of a pendulum to solve this problem:

We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.

We can write this formula for the pendulum when it is on the student's table and when it is falling:

1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:

Canceling out the constants and rearranging, we get:

We know g1; it's simply 10. However, we need to calculate g2, which is the rate at which the pendulum and balloon are accelerating toward the ground. We are given enough information to use the following formula to determine this:

Removing initial velocity and rearranging for acceleration, we get:

Plugging in our values:

This is our g2. We now have all of the values to solve for T2:

We can use the equation for conservation of energy to solve this problem.

If the initial state is when the mass is at its highest position and the final state is when the mass is at its lowest position, then we can eliminate initial kinetic energy and final potential energy:

Substituting expressions in for each term, we get:

Canceling out mass and rearranging for height, we get:

Thinking about a pendulum practically, we can write the height of the mass at any given point as a function of the length and angle of the pendulum:

Think about how this formula is written. The second term gives us how far down the mass is from the top point. Therefore, we need to subtract this from the length of the pendulum to get how high above the lowest point (the height) the mass currently is.

Substituting this into the previous equation, we get:

Rearrange to solve for the angle:

We have values for each variable, allowing us to solve:

The mass of a pendulum has no effect on its period. The equation for the period of a pendulum is

, which does not include mass.

The period and frequency of a pendulum depend only on its length and the gravity force constant, . Changing the mass of the pendulum does not affect the frequency, and since the student released the new pendulum from the same displacement as the old, the amplitude and phase remain the same, and the equation of motion is the same for both pendula.

The period of a pendulum is given by the following formula:

Substituting our values, we obtain:

Roughly 6.3 seconds is the time it takes for the pendulum to complete one period.

The period of a simple pendulum is given by:

Where is the period of the pendulum, is the length of the pendulum, and is the gravitational constant of the planet we are on. Thus on Earth, the period is given by:

With being Earth's gravitational constant. The period on the Moon is given by:

With being the Moon's gravitational constant. Since the Moon's gravity is 6 times weaker than that of Earth's, we have:

Plug this value into the Moon pendulum equation:

Since ,

Substituting this into the above expression gives us

The frequency of a pendulum is given by:

Where is the length of the pendulum and is the gravity constant. Notice how the frequency is independent of mass.

Plugging in values: