Uniform circular motion

Uniform circular motion refers to the movement of an object along a circular path with constant speed. Despite the constant speed, the direction of the object's velocity changes continuously, which means the object is undergoing acceleration.

Velocity in uniform circular motion

Using the equation for average speed, we can derive an equation for the speed of an object in uniform circular motion:

$| \vec {v} | = \frac {d} {t} = \frac {2 \pi r} {T} = 2 \pi r f$

Where $| \vec {v} |$ is the speed, or magnitude of the object in circular motion, $2 \pi r$ is the circumference of the circle, $T$ is the period of the motion (the time it takes to complete one revolution), and $f$ is the frequency of the duration (Note: $T = \frac {1} {f}$).

In terms of the direction of the velocity of the object, it is always perpendicular to the centripetal acceleration.

Centripetal Acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path at constant speed, directed towards the center of the circle. It is given by:

$\vec {a}_c = \frac{v^2}{r}$

Where $\vec {a}_c$ is the centripetal acceleration, $v$ is the velocity of the object, and $r$ is the radius of the circular path.

Centripetal Force

Centripetal force is the force required to keep an object moving in a circular path and is directed towards the center of the circle. It can be calculated using the formula:

$\vec {F}_c = m \vec {a}_c = \frac{mv^2}{r}$

Where $\vec {F}_c$ is the centripetal force, $m$ is the mass of the object, $v$ is the velocity, and $r$ is the radius of the circular path.

Alternate formula

By substituting the equation of velocity in circular motion into the centripetal force, you obtain the following equation:

$\vec {F}_c = \frac {mv^2} {r} = \frac {m ( \frac {4 \pi^2 r^2} {T^2} )} {r} = \frac {4 \pi^2 r m} {T^2}$

Example

A $1000kg$ car is driving on a highway and turns at a curve with a radius of $100m$ at a speed of $25.0 \frac {m} {s}$.

a) Calculate the centripetal acceleration applied onto the car.

b) Calculate the centripetal force applied onto the car.

Answer

a) The centripetal acceleration can be calculated by using its formula:

$\vec {a}_c = \frac {v^2} {r} = \frac {(25.0 \frac {m} {s})^2} { 100m } = 6.25 \frac {m} {s^2}$

b) We can determine the centripetal force by using one of its formulas:

$\vec {F}_c = m \vec {a}_c = (1000kg)(6.25 \frac {m} {s^2}) = 6250N$

Vertical circular motion

When an object is moving in a circular motion vertically, it will have a different centripetal force at each position. The force of gravity $\vec {F}_g$ always points down, while the force of tension from the rope $\vec {F}_T$ always points to the center of the circle.

At the top of circle, both the force of tension and gravity point downwards. At the bottom of the circle, the tension force points upwards, while the force of gravity points down.

$\vec {F}_{c_{\text{top}}} = \vec {F}_T + \vec {F}_g \quad \text{and} \quad \vec {F}_{c_{\text{bottom}}} = \vec {F}_T - \vec {F}_g$

Please note that the centripetal force $\vec {F}_c$ is the net force in the system, not to be confused by the tension force.

Minimum speed to prevent object from falling at the top

You may encounter problems asking you to determine the minimum speed required to keep the ball from falling down at the top of the circle.

In these scenarios, the force of tension $\vec {F}_T = 0$, so:

$\vec {F}_c = \vec {F}_g \quad \Rightarrow \quad \frac {m v^2} {r} = mg \quad \Rightarrow \quad v = \sqrt {rg}$

Example

How fast would an child need to be on going a swing in order to make a full revolution on a swing with a radius of $2.00m$?

Answer

By using the equation $v = \sqrt {rg}$, we get:

$v = \sqrt {rg} = \sqrt { (2.00m)(9.81 \frac {m}{s^2}) } = 19.62 \frac {m} {s}$

Example

A string can withstand a maximum force of $100N$ when spun in a vertical circle. If the string is $1.00m$ long, and is spinning a $3.00kg$ mass, what is the maximum speed that it can spin the mass without breaking?

Answer

At the bottom of the spin, the force of gravity is the largest, so we should look at the maximum speed at the bottom.

To determine the maximum speed, assume the tension of the string is exerting a force of $100N$ upwards:

$\vec {F}_c = \vec {F}_T - \vec {F}_g = (100N) - (3.00kg)(9.81 \frac {m} {s^2}) = 70.6N$

Now, use the equation of the centripetal force that includes the velocity:

$\quad \vec {F}_c = \frac {m v_{ \text {max} }^2} {r}$

$v_{ \text {max}} = \sqrt { \frac { \vec {F}_c r } {m} }$

$v_{ \text {max}} = \sqrt { \frac { (70.6N)(1.00m) } {(3.00kg)} }$

$v_{ \text {max}} = 4.85 \frac {m} {s}$

Example

A $5.00kg$ mass is attached to a strong $2.00m$ rope, and it spins in a vertical circular motion. The mass swings every $1s$.

a) What is the tension in the rope at the top?

b) What is the tension in the rope at the bottom?

Answer

a) We can determine the centripetal force using the second equation we learned:

$\vec {F}_c = \frac {4 \pi^2 r m} {T^2} = \frac {4 \pi^2 (2.00m)(5.00kg)} {(1s)^2} = 395N$

Now we must determine the force of gravity on the mass:

$\vec {F}_g = m \vec {g} = (5.00kg)(9.81 \frac {m} {s^2}) = 46.1N$

Since the mass is at the top, we need to use the equation $\vec {F}_c = \vec {F}_T + \vec {F}_g$ since both the gravity and the tension point downwards:

$\vec {F}_c = \vec {F}_T + \vec {F}_g \ \Rightarrow \ \vec {F}_T = \vec {F}_c - \vec {F}_g = 395N - 46.1N = 349N$

$\therefore$ The tension on the rope at the top is $349N$.

b) We already know the centripetal force and gravity applied onto the mass, so all we need to do is use the equation for the bottom of the circular motion:

$\vec {F}_c = \vec {F}_T - \vec {F}_g \ \Rightarrow \ \vec {F}_T = \vec {F}_c + \vec {F}_g = 395N + 46.1N = 441N$

$\therefore$ The tension on the rope at the bottom is $441N$.

Newton's law of universal gravitation

Newton's law of universal gravitation states that the gravitational force between to objects is proportional on the product of both of their masses, and inversely proportional to one over the distance between squred.

$\vec {F}_g \propto m_1 m_1 \quad \text {and} \quad \vec {F}_g \propto \frac {1} {r^2}$

This law also describes an equation for the gravitational force attracting them:

$\vec {F}_g = \frac {G m_1 m_2} {r^2}$

Where $\vec {F}_g$ is the gravitational force acting on both objects, $m_1$ and $m_2$ are the masses of both objects, $r$ is the distance between them, and $G$ is the gravitational constant.

The gravitational constant is a value obtained experimentally and has a value of $G = 6.67 \times 10^{-11} \frac {N \cdot m^2} {kg^2}$

Satellites

A satellite orbiting earth does not fall into the earth because of its velocity constantly moving it away, but it also does not drift away due to its gravitational pull. Due to this, there is a feeling of weightlessness when inside it.

A satellite is formed when the centripetal force is equal to the force of gravity applied onto the object:

$\vec {F}_c = \vec {F}_g \quad \Rightarrow \quad \frac {m_s v^2} {r} = \frac {G m_E m_s} {r^2} \quad \Rightarrow \quad v = \sqrt { \frac {G m_E} {r} }$

Where $m_E$ is the mass of Earth, $m_s$ is the mass of the satellite and $r$ is the radius of the satellite's orbit with Earth.

Period formula

We can also determine the time it takes the satellite to go around the Earth once (it's period):

$v = \frac {d} {\Delta t} = \frac {2 \pi r} {T} \quad \Rightarrow \quad T = \frac {2 \pi r} {v}$

We can also use the equation $v = \sqrt { \frac {G m_E} {r} }$ to come up with another equation:

$T = \frac {2 \pi r^{ \frac {3} {2} }} { \sqrt {G m_E } }$

Example

Calculate the gravitational force between two objects that are $12.0m$ apart. One object is $100kg$ and the other is $150kg$.

Answer

$\vec {F}_g = \frac {G m_1 m_2} {r^2} = \frac {(6.67 \times 10^{-11} \frac {N \cdot m^2} {kg^2})(100kg)(150kg)} {(12.0m)^2} = 6.95 \times 10^{-9} N$

Example

A satellite completes one revolution around the earth geosynchronously ($1 \text {day}$). Determine the height of the satellite if the earth's sea level is $6.38 \times 10^6 m$ and its mass is approximately $5.98 \times 10^{24} kg$.

Answer

First let's calculate the period in terms of seconds:

$T = 1 \ \text {day} \times 24 \frac { \text {hours} } { \text {day} } \times 60 \frac { \text {minutes} } { \text {hour} } \times 60 \frac { \text {seconds} } { \text {minute} } = 86400s$

Now, we need to re-arrange the equation for the period of a satellite to obtain the height from the center of Earth:

$\, T = \frac {2 \pi r^{ \frac {3} {2} }} { \sqrt {G m_E } }$

$r^{ \frac {3} {2} } = \frac { T \sqrt {G m_E } } {2 \pi }$

$\ \, \, r = \sqrt [ \frac {2} {3} ] {\frac { T \sqrt {G m_E } } {2 \pi }}$

$\ \, \, r = \sqrt [ \frac {2} {3} ] {\frac { (86400) \sqrt {(6.67 \times 10^{-11} \frac {N \cdot m^2} {kg^2})(5.98 \times 10^{24} kg) } } {2 \pi }}$

$\ \, \, r = 4.225 \times 10^7 m$

Now, we need to determine the height of the satellite above sea level:

$\text {height} = r - r_{ \text{sea} } = (4.225 \times 10^7 m) - (6.38 \times 10^6 m) = 3.59 \times 10^7 m$

$\therefore$ The satellite is $3.59 \times 10^7 m$ above sea level.