Notes For:

Linear Vs. Circular MotionEdit

  • Linear motion is described by position, velocity, and acceleration.
  • Circular motion repeats itself in circles around the axis of rotation.
    • Planets in orbit
    • Children on a merry-go-round
    • Spinning a basketball

Rotating versus RevolvingEdit

  • There is an imaginary line called an axis that either runs through the center of an object (internal) of is outside of the object itself (external).
    • A basket ball spinning on your finger is rotating about its own axis.
    • A child on a merry-go-round is revolving around the outside axis of the merry-go-round
    • The Earth rotates daily around its own internal axis and we experience day and night.
    • The Earth revolves ounce a year around the external axis of the sun.

Angular PositionEdit

  • \Theta – angle measured from a reference line.
    • The reference line simply defines \Theta = 0 and is analogous to the origin in the linear coordinate system.
  • The reference line begins at the axis of rotation. It can point in any direction, but must be used consistently thereafter.
  • \Theta > 0 (Positive) has counterclockwise rotation from the reference line.
  • \Theta < 0 (Negative) has clockwise rotation from reference line.

Units for Measuring AnglesEdit

  • Degrees (º)
  • Revolutions (1 complete turn around a circle, corresponds to 360º)
    • 1 revolution = 360º
  • Radians
    • A radian is approximately 57.3º
    • Radians are better than degrees because the radian is a ratio of 2 lengths
      • Arc Length
      • Radius
    • The radian does not have any unit
  • A radian is the angle for which the arc length (part of the circumference) on a circle of radius r is equal to the radius of the circle
    • So, for an angle of 1 radian, arc length (s) = radius (r)
    • s = r

Solving for Arc LengthEdit

To Solve for Arc Length
s = r \Theta

  • s = Arc Length
  • r = Radius
  • \Theta = Angle

To solve for \Theta
\Theta = \frac{s}{r}

  • \Theta = Angle
  • s = Arc Length
  • r = Radius

This relation ONLY works for radians, not degrees or revolutions.


  • One complete rvolution is total circumference of the circle.
    • C = 2 \pi r
  • Failed to parse (lexing error): 1 \text{Revolution} = 360º = 2 \pi
  • Failed to parse (lexing error): \Theta = \frac{\pi \Theta}{180º}

Angular DisplacementEdit

  • Angular displacement describes how many an object has rotated.
  • Angular displacement is the change in arc length divided by the radius.
  • \Delta \Theta = \frac {\Delta s}{r}

Sample ProblemEdit

While traveling on a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the child's angular displacement is 165º, what is the radius of the carousel?

  • s = -11.5 m
  • \Theta = -165 º
  • r = ?
  • \Theta = \frac{s}{r}
  • r = \frac{s}{\Theta}
  • Failed to parse (lexing error): \Theta = \frac {\pi \Theta}{180º}
  • Failed to parse (lexing error): \Theta = \frac {-165º \pi}{180º} = -2.88
  • r = \frac{-11.5m}{-2.88} = 3.99 m

Angular SpeedEdit

  • Angular speed is the rate at which an object rotates or revolves.
    • Also called rotational speed.
    • Describes the amount an object turns in a specific time period.
    • You calculate it by dividing the number of turns by the time it takes.
  • Angular Speed \left ( \omega \right ) equals the change in angle divided by the change in time.

\omega = \frac{\Theta}{t}

  • \omega – Angular Speed
  • \Theta – Angel (in Radians)
  • t – time

  • Angular speed is similar to linear speed except angle replaces distance.

Unit of Angular SpeedEdit

  • Rotation per Minute (RPM)
  • Change in angle per unit of time
    • Degrees per second \left ( \frac{\circ}{s} \right )
    • Radians per second \left ( \frac{radians}{s} \right )
  • The radian is more convenient unite to calculate angular speed than degrees.

Practice ProblemEdit

A bicycle wheel makes 6 turns in 2 seconds. What is its angular speed in radians per second?

  • 6 turns = 6 revolutions
  • t = 2 seconds
  • Conversion: 6 \text{Rev} = 12 \pi
  • \omega = \frac{\Theta}{t} = \frac{12 \pi}{2 s} = 6 \pi \frac{\text{rad}}{s} = 18.8 \frac{\text{rad}}{s}

Angular AccelerationEdit

  • Average angular acceleration of an object, \alpha_\text{avg}, is the change of its angular speed, \Delta \omega, in a certain time interval.

\alpha_\text{avg} = \frac {\Delta \omega}{\Delta t}

  • \alpha_\text{avg} – Average Angular Acceleration \left ( \frac{\text{rad}}{s^2} \right )
  • \Delta \omega – Change in angular speed
  • \Delta t – Change in time

\alpha_\text{avg} = \frac {\omega_2 - \omega_1}{t_2 - t_2}

  • \alpha_\text{avg} – Average Angular Acceleration \left ( \frac{\text{rad}}{s^2} \right )
  • \omega_2 – Angular Speed 2 \left ( \frac{m}{s} \right )
  • \omega_1 – Angular Speed 1 \left ( \frac{m}{s} \right )
  • t_2 – Time 2 (s)
  • t_1 – Time 1 (s)

Angular Acceleration and SpeedEdit

  • A point on the rim of a rigid rotating object has the same angular speed and the same angular acceleration as a similar point nearer to the center.
Rotational motion with constant angular acceleration. Item Linear motion with constant acceleration
\Theta Distance x
\omega Avg. Speed v
\alpha Avg. Acceleration a
\omega_f = \omega_i + \alpha \Delta t Find Final Speed v_f = v_i + a \Delta t
\Delta \Theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2 Find Change in Distance \Delta x = v_i \Delta t + \frac{1}{2} a (\Delta t)^2
\omega_f^2 = \omega_i^2 + 2 \alpha(\Delta \Theta) Find Final Speed v_f^2 = v_f^2 = 2 a (\Delta t)
\Delta \Theta = \frac{1}{2}(\omega_i + \omega_f) \Delta t Find Change in Distance \Delta x = \frac{1}{2} (v_i + v_f) \Delta t

Angular and Linear QuantitiesEdit

  • Almost every linear quantity that we've studied has a corresponding twin in rotational motion.


The wheel of an upside-down bicycle moves through 11.0 rad in 2.0 s. What is the wheel's angular acceleration if its initial angular speed is 2.0 rad/s

  • \Theta = 11.0 rads
  • t = 2 s
  • \omega_i = 2.0 rads/s
  • \alpha = \frac{2 \Theta - 2(\omega_i \Delta t}{t^2}
  • Failed to parse (lexing error): \alpha = \frac{2 11.0 rads - 2(2.0 rads • 2 s}{2 s^2} = 3.5 \frac{rad}{s^2}


Think about a carousel totaling about its center. Any two horses have the same angular speed and singular acceleration regardless of where they are on the carousel. But at different distances from the axis of rotation, they have different tangential speeds. The tangential speed of any point rotating about an axis is also called the instantaneous linear speed of the point. The tangential speed of a horse on the carousel is its speed along a line drawn tangent to its circular path. The horse on the outside must travel the same angular displacement during the same amount of time as the horse on the inside. An object that is farther from the axis of a rigid rotating body must travel at a higher tangential speed around the circular path.

Linear vs. Angular SpeedEdit

  • A wheel rolling along the ground has both a linear speed and an angular speed.
  • If the wheel doesn't slip, the 2 speeds are related by the circumference of the wheel.
  • Circumference is the distance a circle and depends on the radius of the circle.
  • Circumference = 2 \pi r
  • A point at the edge of a wheel moves 1 circumference in each turn of the circle.
  • Linear speed (v) is the circumference divided by the time it takes to make 1 turn.
  • If we're working with circumference, angular speed, \omega = \frac{2 \pi}{t}
  • Linear speed = angular speed x radius

v = \omega r

  • v = velocity
  • \omega = angular speed
  • r = radius

Angular speed vs. Linear speedEdit

  • Angular speed of a wheel is the same for all points on the wheel.
  • Linear speed of a point on a wheel depends on how far the point from the center of rotation.
  • Points at the outer edge of a wheel move faster than points nearer to the center.
  • The point at the very center of the wheel (r=0) doesn't move at all.
  • linear speed is perpendicular

Practice ProblemEdit

1) Two children are spinning around on a merry-go-round. Joe is standing 4 meters from the axis of rotation and Sue is standing 2 meters from the axis. Calculate each child's linear speed when the angular speed is 1 rad/s.

  • \omega = 1 rad/s
  • r_j = 4
  • r_s = 2
  • v_j = r_j  \omega = 4 x 1 = 4 \frac{m}{s}
  • v_s = r_s  \omega = 2 x 1 = 2 \frac{m}{s}

2)The radius of a CD in a computer is .0600 m. If a microde riding on the disc's rim geas a tangential speed of 1.88 m/s, what is the disc's angular speed?

  • r = 0.0600 m
  • v_t = 1.88 m/s
  • \omega = ?
  • \omega = \frac{1.88 \frac{m}{s}}{0.0600 m} = 31.3 \frac{rad}{s}

3)A bicycle has wheels that are 70 cm in diameter. The bicycle is moving forward with a linear speed of 11 m/s. Assume the wheels are not slipping! Calculate the angular speed of the wheels in RPM.

  • Diameter = 70 cm
  • \omega = ?
  • v = 11 m/s
  • r = 35 cm
  • \omega = \frac{v}{r} = \frac{11}{35} = 31.4 rad/s
  • 31.4 \frac{rads}{s} \times \frac{60s}{1min} \times \frac{rotation}{2 \pi rad} = 300 RPM

Tangential AccelerationEdit

  • Tangential Acceleration – the instantaneous linear acceleration of an object directed along the tangent to the object's circular path.
  • If a rotating object speeds up, it has angular acceleration.
  • The linear acceleration related to this angular acceleration is the tangential acceleration.

a_t = r \alpha

  • a_t = Tangential Acceleration (\frac{m}{s^2}
  • r = radius
  • \alpha = angular acceleration ({r}{s^2})

Practice ProblemEdit

  • \alpha = 50 {r}{s^2}
  • a_t = 3.3 {m}{s^2}
  • r = \frac{a_t}{\alpha} = \frac{3.3}{50} = 0.66 rads

Centripetal ForceEdit

  • A force is required for any object in circular motion to keep the object moving in a circular motion.
  • This force always points toward the center of the circle.
  • This force is the centripetal force.
  • Any type of force can be a centripetal force if its action causes an object to move in a circle.
    • Friction between a car's tires and the road is the centripetal force that allows the car to turn around a curve.
  • Whether a force makes an object accelerate by changing its speed or by changing its direction depends on the direction of the force.
  • A force applied perpendicular to the motion causes the object to change its path from a line to a circle, without changing speed.
  • Centripetal force is always perpendicular to the object's motion, toward the center of the circle.

Calculating Centripetal ForceEdit

  • The magnitude of the centripetal force needed to move an object in a circle depends on the object's mass and speed, as well as on the radius of the circle.
  • Newton's Second Law – F = ma
    • Newton's 2nd law – the greater the mass of an object, the greater the centripetal force needed to change its motion.
    • As speed increases, a greater centripetal force is required to keep bending an object's path into a circle.
  • The strength of the centripetal force is proportional to the square of the speed.
  • The larger the radius of the circle, the more gradually an object's direction of motion changes
  • Radius and centripetal force are inversely related.

F_c = \frac{m \times v_t^2}{r}

  • F_c – Centripetal Force (N)
  • m – Mass (kg)
  • v_t – Linear Speed (\frac{m}{s})
  • r – Radius (m)

F_c = m \times r \times \omega^2

  • F_c – Centripetal Force (N)
  • m – Mass (Kg)
  • r – Radius (m)
  • \omega – Angular Speed (\frac{\text{radians}}{s})

Practice ProblemEdit

A 50 kilogram passenger on an amusement park ride stands with his back against the wall of a cylindrical room with radius of 3 meters. What is the centripetal force of the wall pressing into his back when the room spins and his is moving at 6 m/s?

  • F_c – ? N
  • m – 50 kg
  • v_t – 6 \frac{m}{s}
  • r – 3 m

F_c = \frac{50 \times 6^2}{3} = 600 N

Centripetal AccelerationEdit

  • Centripetal acceleration is acceleration directed toward the center of a circular path due to the centripetal force.
  • THe direction of an object's acceleration is always in teh same direction as the net force.
  • Centripetal acceleration is in the same direction as the centripal force, toward the center of teh circle and is constantly changing

Finding Centripetal AccelerationEdit

  • According to Newton's Second Law: F = ma

a_c = \frac{v_t^2}{r}

  • a_c – Centripetal Acceleration (\frac{m}{s^2})
  • v_t – Linear Speed (\frac{m}{s})
  • r – Radius (m)

a_c = r \times \omega^2

  • a_c – Centripetal Acceleration (\frac{m}{s^2})
  • r – Radius (m)
  • \omega – Angular Speed (\frac{\text{radians}}{s})

Practice ProblemEdit

A motorcycle drives around a bend with a 50 meter radius at 10 m/s. Find the motorcycle's centripetal acceleration and compare it with g, the acceleration of gravity.

  • a_c = ?
  • v_t – 10 m/s \frac{m}{s}
  • r – 50 m

a_c = \frac{10 \frac{m}{s}^2}{10} = 2 \frac{m}{s^2}

Gravitational ForceEdit

  • Most of the planets move in nearly circular orbits around the sun.
  • The gravitational force keeps the planets from wandering off in a straight line
  • The gravitational force is a field force that acts at a distance between two masses regardless of the medium that separates them.
  • Gravitational force exists between any two masses regardless of size or composition.
  • Gravity is an attractive force.
  • Gravity obeys Newton's third law.
    • F_\text{mE} = -F_\text{Em}

Newton's Law of Universal GravitationEdit

F_g = G \times \frac{m_1 \times m_2}{r^2}

  • G = Gravitational Constant
  • m_1 = Mass one
  • m_2 = Mass two
  • r = Distance between two masses

G = 6.673 \times 10^\text{-11} \frac{N \times m^2}{kg^2}

  • Inverse-square law: force between 2 masses decreases as the distance between them increase.

Practice ProblemEdit

F_g = G \times \frac{m_1 \times m_2}{r^2}

  • G = 6.673 \times 10^\text{-11} \frac{N \times m^2}{kg^2}
  • m_1 = 7.36 \times 10^ \text{22} kg
  • m_2 = 5.98 \times 10^ \text{24} kg
  • r = 3.84  \times 10^ \text{8} m

F_g = 6.673 \times \frac{10^\text{-11} \frac{N \times m^2}{kg^2} \times 7.36 \times 10^ \text{22} kg \times 5.98 \times 10^ \text{24} kg}{3.84  \times 10^ \text{8} m^2} = 1.99 \times 10^\text{20} N

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