Physics Study Guide for the HESI Exam

Rotation

Linear motion is when an object moves in a straight line, whereas rotational motion involves an object rotating around a fixed point or spinning on an axis, like the Earth rotating around the Sun. Rotational motion is involved in everything from spinning tires to blood-spinning in a centrifuge.

Angular Displacement

Angular displacement is the angle through which an object rotates. It’s measured in degrees or radians. It tells us how far an object has “turned” around a fixed axis. For example, when the hands of the clock move from $12$ to $3$, the minute hand has an angular displacement of $90^\circ$.

The formula for angular displacement is:

\[\theta = \omega \cdot t\]

where $\theta$ is the angular displacement in radians, $\omega$ is the angular (or rotational) velocity in radians per second, and $t$ is the time in seconds.

Note: You should know how to convert between radians and degrees. These two equations are helpful:

\[360^\circ = 2 \pi \text{ rad}\] \[1 \text{ rad} = \frac{180^\circ}{\pi} \approx 57.296^\circ\]

Let’s try an example problem that involves angular displacement.

A car runs for $12$ seconds around a circular track with an angular velocity of $\frac{\pi}{4}$ radians per second. What will be its angular displacement in degrees?

Solution

This is a two-step problem. First, you need to find the angular displacement in radians:

\[\theta = \omega \cdot t = (\frac{\pi}{4} \text{ rad/s})(12 \text{ s}) = 3 \pi \text{ rad}\]

Now, convert from radians to degrees:

\[3 \pi \text{ rad} \cdot \frac{360^{\circ}}{2 \pi \text{ rad}} = 3 \cdot 180 = 540^{\circ}\]

Thus, the angular displacement is $540^{\circ}$.

Speed

Speed tells us how fast something is moving. When an object is rotating, we deal with how quickly it spins or turns over time.

Average Speed

In rotational motion, the average speed is not how fast an object has traveled along a circular path (that’s rotational velocity), but how far the object has physically moved around the circle. So instead of using regular distance, we use arc length ($l$), which is the distance along the curved path the object follows:

\[\overline{s} = \frac{l}{t}\]

Now, if we need to find arc length, we use this formula:

\[l = r \theta\]

where $l$ is the arc length in meters, $r$ is the radius of the circle in meters, and $\theta$ is the angular displacement in radians.

Let’s try an example problem.

A point on the edge of a spinning wheel travels through an angular displacement of $3 \pi$ radians in $6$ seconds. If the radius of the wheel is $0.5$ meters, what is the average speed?

Solution

First, find the arc length:

\[l = r \theta = 0.5 \times 3 \pi = 1.5 \pi \text{ m} \approx 4.71 \text{ m}\]

Now you can find the point’s average speed:

\[\overline{s} = \frac{4.71 \text{ m}}{6 \text{ s}} = 0.785 \text{ m/s}\]

Angular Speed

Rotational velocity, also known as angular velocity ($\omega$, discussed earlier), represents how fast an object has traveled along a circular path. It is given by the rotational displacement ($\theta$) in a given time:

\[\omega = \frac{\theta}{t}\]

In problems where you are first given the linear speed ($v$) and radius ($r$), you can also find $\omega$ by relating both motions:

\[v = r \cdot \omega\]

We can rearrange this as such:

\[\omega = \frac{v}{r}\]

Rotational displacement, also known as angular displacement ($\theta$, discussed earlier), is an angle, so the units are degrees or radians. One complete revolution corresponds to an angular displacement of $360$ degrees or $2\pi$ radians.

Here is an example problem involving angular speed.

A disk turns $4 \pi$ radians in $8$ seconds. What is the angular speed?

Solution

Finding the answer is simply a matter of inserting the given values into the formula:

\[\omega = \frac{\theta}{t} = \frac{4 \pi}{8 \text{ s}} = 1.57 \text{ rad/s}\]

Angular Acceleration

Angular acceleration is how quickly angular speed changes. It is measured in radians per second squared ($\text{rad/s}^2$).

The formula for angular acceleration is:

\[\alpha = \frac{\Delta \omega}{t}\]

where $\alpha$ is the angular acceleration, $\Delta \omega$ is the change in angular velocity (the final rate minus the initial rate), and $t$ is the time.

Let’s try an example.

A wheel speeds up from $2 \text{ rad/s}$ to $6 \text{ rad/s}$ in $2 \text{ s}$. What is its angular acceleration?

Solution

Again, we are given the values, and we just need to insert them into the formula:

\[\alpha = \frac{\Delta \omega}{t} = \frac{6 \text{ rad/s} - 2 \text{ rad/s}}{2 \text{ s}} = \frac{4 \text{ rad/s}}{2 \text{ s}} = 2 \text{ rad/s}^2\]

Uniform Circular Motion

Uniform circular motion is when an object moves in a circular path at a constant speed. Even though the speed stays the same, the velocity is constantly changing because the direction of motion is always changing.

Remember that velocity includes both speed and direction, so changing direction means changing velocity.

Centripetal Acceleration

Centripetal acceleration is the acceleration directed toward the center of the circular path. It is a consequence of the centripetal force (defined below) and describes how quickly the object’s velocity is changing direction.

The formula to calculate centripetal acceleration ($a_c$) is:

\[a_c = \frac{v^2}{r}\]

where $v$ is the linear speed and $r$ is the radius of the circular path.

Centripetal Force

Centripetal force is the net force that keeps an object moving in a circular path. It is always directed inward, toward the center of the circle. Centripetal force is what keeps roller coaster riders in their seat even when upside down.

Just as force is mass times acceleration, centripetal force is mass times centripetal acceleration. Therefore, the formula to calculate centripetal force ($F_c$) is:

\[F_c = \frac{mv^2}{r}\]

where $m$ is the mass of the object, $v$ is the linear speed, and $r$ is the radius of the circle.

Energy

Energy is the ability to do work or cause change. It shows up in many forms like heat, light, motion, and sound. In physics, we focus on mechanical energy, which includes kinetic energy (motion) and potential energy (position or condition). Energy, which is measured in joules (J), is conserved, meaning that it can’t be created or destroyed, only transformed from one form to another.

Kinetic Energy

Kinetic energy is the energy of an object due to its motion. If an object is moving, whether it’s a baseball flying through the air or you walking down the hallway, it has kinetic energy. Determining how much kinetic energy ($KE$) requires knowing the object’s mass ($m$) and velocity ($v$):

\[KE = \frac{1}{2}mv^2\]

Let’s do an example problem to see how this works in the real world.

A $2\text{ kg}$ ball is rolling at $3\text{ m/s}$. What is its kinetic energy?

Solution

Just insert the given values into the formula:

\[KE = \frac{1}{2} mv^2 = \frac{1}{2} (2 \text{ kg})(3 \text{ m/s})^2 = 9 \text{ J}\]

Therefore, while the ball is rolling, its kinetic energy is $9$ joules.

Potential Energy

Potential energy is the energy of an object due to its position within a field. For example, when that two-kilogram ball is resting on top of a hill, it has potential energy, as it has mass and can roll down that hill. The gravitational potential energy ($PE$) of an object is found with the following formula:

\[PE = mgh\]

where $m$ is the mass, $g$ is the gravitational field strength (a constant of $9.8 \text{ m/s}^2$), and $h$ is the height of an object.

The total energy of a system ($U$) is constant and is equal to the sum of the kinetic and potential energy:

\[U = KE + PE\]

As the kinetic energy increases, the potential energy must decrease by the same amount to keep $U$ constant. For example, when the ball is dropped, its $PE$ is converted to $KE$.

Joules

Joules (J) are the units used to measure energy. One joule is the energy needed to move a one-kilogram object by one meter using a force of one newton.

Let’s try an example problem that involves calculating joules.

How much energy is needed to lift a five-kilogram bowling ball two meters high?

Solution

The energy needed to lift an object against gravity is equal to its gravitational potential energy ($PE$), which is calculated using the formula $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height. Substituting these values into the equation, we get:

\[PE = (5 \text{ kg})(9.8 \text{ m/s}^2)(2 \text{ m}) = 98 \text{ J}\]

So, $98$ joules of energy are required to lift the bowling ball. In other words, the ball’s potential energy after being lifted is $98$ joules. Energy was converted to potential energy and transferred from the person lifting the bowling ball to the bowling ball in order to lift it.

Momentum and Impulse

Momentum and impulse are closely related. Momentum is about motion, and impulse is about changing that motion. This is all related to linear motion (motion in a straight line).

Momentum

Momentum is how much motion something has and depends on the object’s mass and velocity. It is calculated using the following equation:

\[p = m \cdot v\]

where $p$ is the momentum, in units of newton-seconds ($\text{N}\text{·}\text{s}$) or kilogram-meters per second ($\text{kg}\text{·}\text{m/s}$), $m$ is the mass, and $v$ is the velocity.

In a closed system, the total momentum remains constant over time, which is known as the principle of conservation of momentum. This means that for two objects that collide:

\[m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f}\]

where the $i$ indicates initial momentum before the collision and $f$ indicates final momentum after the collision.

The Impulse Equation

Impulse is the change in momentum caused by a force over time. It is calculated in newton-seconds and is defined by the following formula:

\[I = F \cdot t\]

where $I$ is the impulse, $F$ is the force, and $t$ is the time.

As was already said, impulse is the change in momentum ($\Delta p$), so we can also think of it as:

\[I = \Delta p = m \cdot \Delta v\]

where $m$ is mass and $\Delta v$ is the change in velocity.

An example problem will help make this concept clearer.

A $four-kilogram object speeds up from two to six meters per second in two seconds. What was the impulse?

First, calculate the change in velocity:

\[\Delta v = 6 \text{ m/s} - 2 \text{ m/s} = 4 \text{ m/s}\]

Then, calculate the impulse:

\[I = m \Delta v = 4 \text{ kg} \times 4 \text{ m/s} = 16 \text{ N}\text{·}\text{s}\]

Universal Gravitation

Universal gravitation is the force of attraction between any two objects with mass. This gravitational force is stronger when the objects are heavier as well as when they are closer together. The law of universal gravitation was first described by Sir Isaac Newton.

The Law of Universal Gravitation

The force of gravity ($F_g$) is the attractive force that is felt between all objects. The gravitational force felt by two objects is directly proportional to the product of their masses ($m_1$ and $m_2$) and inversely proportional to the square of the distance ($r$) between them. This means, the more massive the objects and the closer together they are, the stronger the gravitational force produced. This is described by Newton’s law of gravitation, represented by this formula:

\[F_g = G\frac{m_1m_2}{r^2}\]

where $G$ is the gravitational constant $6.674\times10^{-11} 2\, \frac{\text{m}^{3}}{\text{kg} \cdot \text{s}^{2}}$.

Let’s do an example problem. This can be complicated, but we’ll walk you through each step, including how to handle the exponents.

What is the gravitational force between a $70 \text{ kg}$ person and the Earth (mass = $5.972 \times 10^{24} \text{ kg}$), assuming they are $6.37 \times 10^6 \text{ m}$ apart (Earth’s radius)?

Solution

To solve this problem we need to recognize that the gravitational force depends on both masses and the distance between their centers of mass. Here, the distance given is from the person’s center of mass to the Earth’s center, so we’ll use the Earth’s radius.

We will need to use Newton’s law of gravitation:

\[F_g = G\frac{m_1m_2}{r^2}\]

Substituting the known values, we get:

\[F_g = (6.673\times10^{-11})\frac{(70)(5.972 \times 10^{24})}{(6.37 \times 10^6)^2}\]

Now, we handle the powers of ten carefully. Multiply the masses first:

\[70 \times 5.972 \times 10^{24} = 418.04 \times 10^{24} = 4.1804 \times 10^{26}\]

Notice that the exponent increased by two because we needed to move the decimal place over two places to be in scientific notation.

Next, square the distance:

\[(6.37\times10^6)^2 = 40.58\times10^{12}= 4.058\times10^{13}\]

Now, we will divide the numerator by the denominator. Remember, with division, the exponents subtract:

\[\frac{4.1804 \times 10^{26}}{4.058\times10^{13}} = 1.03\times10^{13}\]

Finally, multiply by $G$. In this case, because we are multiplying, the exponents add:

\[F_g = (6.674\times10^{-11})(1.031\times10^{13}) = 6.88\times10^2 \text{ N}\]