When an object is rotating, we measure its motion using radians
I. Radian = arc length that = radius of circle Video p. 1-3
360 degrees - One full rotation of a sphere is radians
90 degrees = radians
Ex 1) How many radians does the arc s represent?
Number of radians $\theta = $
Equation: $\theta = \frac{s}{r}$
And $d = r\theta$
Ex 2) A wheel with a radius of 2.0 m rolls 5.0 m along the floor. Find the number of radians rotated through. $\theta =$
TEACHER CER KEY (AP SCORE: 5/5)
Claim: The wheel rotates through exactly $2.5$ radians.
Evidence: Radius $r = 2.0\text{ m}$, Linear distance traveled (arc length) $s = 5.0\text{ m}$.
Reasoning: By definition, an angular displacement in radians ($\theta$) is the ratio of the arc length traveled ($s$) to the radius of the circular path ($r$). Applying $\theta = s/r$: $\theta = (5.0\text{ m}) / (2.0\text{ m}) = 2.5$. Radians are a dimensionless ratio, thus no linear units remain.
Ex 3) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find its angular velocity.
Angular velocity = radians/sec
Formula: $\omega = \frac{v}{r}$ or $v = r\omega$
TEACHER CER KEY (AP SCORE: 5/5)
Claim: The ball possesses an angular velocity ($\omega$) of $10\text{ rad/s}$.
Evidence: Tangential velocity $v = 5.0\text{ m/s}$, Radius $r = 0.50\text{ m}$.
Reasoning: For an object rolling without slipping, the tangential velocity of the outer edge is equal to the product of the radius and the angular velocity ($v = r\omega$). Rearranging for $\omega$ gives $\omega = v/r$. Substituting the given values: $\omega = (5.0\text{ m/s}) / (0.50\text{ m}) = 10\text{ rad/s}$. This represents the rate of change of the angular position per unit time.
Ex 4) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find a period of rotation.
$T = \frac{2\pi \text{ rad}}{\omega}$
IV. Frequency
Frequency is the number of full rotations completed per second.
$f = \frac{1}{T} = \frac{\omega}{2\pi}$
Ex 5) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find frequency of rotation.
TEACHER CER KEY (AP SCORE: 5/5)
Claim: The period $T$ is $0.628\text{ s}$ and the frequency $f$ is $1.59\text{ Hz}$.
Evidence: Angular velocity $\omega = 10\text{ rad/s}$.
Reasoning: One full rotation corresponds to an angular displacement of $2\pi$ radians. The Period ($T$) is the time required for this displacement: $T = 2\pi/\omega$. Thus, $T = 2\pi/10 \approx 0.628\text{ s}$. Frequency ($f$) is the reciprocal of the period, representing how many full cycles occur in one second: $f = 1/T = 1/0.628 \approx 1.59\text{ cycles/sec (Hz)}$.
Ex 7) A ball with a radius of .10 m starts from rest and accelerates down a 10. m incline and attains a velocity of 8.0 m/s. Find its ANGULAR acceleration.
Ex 6: $\alpha = 1250\text{ rad/s}^2$.
Reasoning: First, calculate linear acceleration $a = \Delta v / \Delta t = (5.0 - 0) / 0.10 = 50\text{ m/s}^2$. Angular acceleration ($\alpha$) is the rate of change of angular velocity, related to linear acceleration by $a = r\alpha$. Converting radius to meters ($r = 0.04\text{ m}$), we solve for $\alpha = a/r = 50 / 0.04 = 1250\text{ rad/s}^2$.
Ex 7: $\alpha = 32\text{ rad/s}^2$.
Reasoning: **Strategy 1 (Linear first):** Use $v_f^2 = v_i^2 + 2ad$ to find $a$: $8.0^2 = 0 + 2(a)(10) \implies 64 = 20a \implies a = 3.2\text{ m/s}^2$. Then $\alpha = a/r = 3.2 / 0.1 = 32\text{ rad/s}^2$. **Strategy 2 (Angular first):** Convert all variables to angular ($v \rightarrow \omega = 80\text{ rad/s}$, $d \rightarrow \theta = 100\text{ rad}$) and solve $\omega_f^2 = \omega_i^2 + 2\alpha\theta$: $80^2 = 0 + 2(\alpha)(100) \implies 6400 = 200\alpha \implies \alpha = 32\text{ rad/s}^2$. Both yield the same result because the relationship between linear and angular motion is constant.
A bicycle tire of radius 0.33 m moves forward, without slipping, for a distance of 0.27 m. How many radians does the wheel rotate?
A bicycle tire of radius 0.33 m moves forward, without slipping, for a distance of 0.27 m. How many degrees does the wheel rotate?
A flat disc rotates $1.5\pi$ rad in 12 s. What is its angular velocity?
An electric train rides a circular track of radius 0.39 m and its wheels rotate an arc length (linear distance) of 0.96 m in 4.0 s. What is the angular velocity of the wheels?
A boy pulls a wagon in a circle. The wagon starts from rest and accelerates to an angular velocity of 0.25 rad/s in 1.2 s. What is the wagon’s angular acceleration?
A car tire rotating with an angular velocity of 42 rad/s accelerates at a constant rate to 51 rad/s in 3.0 s. What is its angular acceleration?
A bicycle has wheels with a radius of 0.25 m. The wheels are moving at a constant angular velocity of 3.1 rad/s. What is the linear velocity of the bicycle?
A motorbike has tires of a radius of 0.25 m. If the motorbike is traveling at 1.7 m/s, what is the angular velocity of the tires (assume the tires are rotating without slipping)?
A cyclist accelerates uniformly from rest to a linear velocity of 1.0 m/s in 0.75 s. The bicycle tires have a radius of 0.38 m.
What is the angular velocity of each bicycle tire at t = 0.75 s?
What is the angular acceleration of each tire?
What is the tangential acceleration of each tire at r = 0.38 m?
What is the centripetal (radial) acceleration of each tire at r = 0.38 m?
What is the total linear acceleration of each tire at t = 0.75 s?
What is the rotational frequency (f) of each tire?
Unlike record players which rotate at a constant angular velocity, the CD-ROM driver in a CD player rotates the CD at different angular velocities so that the optical head maintains a constant tangential velocity relative to the CD. Assume the CD has a diameter of 0.120 m and that the CD needs to move at a tangential velocity of 1.26 m/s relative to the optical head so the data (music) can be read correctly. Video Question 10
What is the angular velocity when the CD-ROM is reading data at the outer edge of the CD? Give the answer in rpm and rad/s.
At the inner edge of the CD, the CD-ROM rotates at 500. rpm. How far from the center of the CD is the inner edge ($v_{tangential} = 1.26 \text{ m/s}$)?
What is the centripetal (radial) acceleration at the outer edge of the CD?
What is the rotational frequency (f) of the CD when it is reading data at its outer edge?
TEACHER CER KEY (AP SCORE: 5/5)
Q1/Q2: $\theta = 0.82\text{ rad}$ and $46.9^\circ$.
Reasoning: $\theta = s/r = 0.27 / 0.33 = 0.818\text{ rad}$. To convert to degrees, multiply by $180/\pi$: $0.818 \times (180/3.14) \approx 46.9^\circ$.
Q9e: $a_{total} \approx 2.95\text{ m/s}^2$.
Reasoning: At any point in time, a rotating object has both tangential acceleration ($a_t = r\alpha = 1.33\text{ m/s}^2$) and centripetal acceleration ($a_c = v^2/r = 2.63\text{ m/s}^2$). Since these vectors are perpendicular, the total linear acceleration is the magnitude of their resultant: $a = \sqrt{1.33^2 + 2.63^2} \approx 2.95\text{ m/s}^2$.
Q10a: $\omega_{outer} = 21\text{ rad/s}$ ($201\text{ rpm}$).
Reasoning: The diameter is $0.120\text{ m}$, so radius $r = 0.060\text{ m}$. Given $v = 1.26\text{ m/s}$, $\omega = v/r = 1.26 / 0.06 = 21\text{ rad/s}$. To convert to rpm: $(21\text{ rad/s}) \times (60\text{ s/min}) / (2\pi\text{ rad/rev}) \approx 201\text{ rpm}$.
A tricycle wheel of radius 0.11 m is at rest and is then accelerated at a rate of 2.3 rad/s$^2$ for a period of 8.6 s. What is the wheel’s final angular velocity?
A tricycle wheel of radius 0.11 m is at rest and is then accelerated at a rate of 2.3 rad/s$^2$ for a period of 8.6 s. What is the wheel’s final linear velocity?
A potter’s wheel is rotating with an angular velocity of $\omega = 3.2$ rad/s. The potter applies a constant force, accelerating the wheel at 0.21 rad/s$^2$. What is the wheel’s angular velocity after 6.4 s?
A potter’s wheel is rotating with an angular velocity of $\omega = 3.2$ rad/s. The potter applies a constant force, accelerating the wheel at 0.21 rad/s$^2$. What is the wheel’s angular displacement after 6.4 s?
A bicycle wheel with a radius of 0.38 m accelerates at a constant rate of 4.8 rad/s$^2$ for 9.2 s from rest. How many revolutions did it make during that time?
A bicycle wheel with a radius of 0.38 m accelerates at a constant rate of 4.8 rad/s$^2$ for 9.2 s from rest. What was its linear displacement during that time?
A Frisbee of radius 0.15 m is accelerating at a constant rate from 7.1 revolutions per second to 9.3 revolutions per second in 6.0 s. What is its angular acceleration?
A Frisbee of radius 0.15 m is accelerating at a constant rate from 7.1 revolutions per second to 9.3 revolutions per second in 6.0 s. What is its angular displacement during that time?
A record is rotating at 33 revolutions per minute. It accelerates uniformly to 78 revolutions per minute with an angular acceleration of 2.0 rad/s$^2$. Through what angular displacement does the record move during this period? Video 9 - 10
What is the angular acceleration of a record that slows uniformly from an angular speed of 45 revolutions per minute to 33 revolutions per minute in 3.1 s?
TEACHER CER KEY (AP SCORE: 5/5)
Q3/Q4: $\omega_f = 4.54\text{ rad/s}$ and $\Delta\theta = 24.8\text{ rad}$.
Reasoning: For constant angular acceleration, we use $\omega_f = \omega_i + \alpha t$. $\omega_f = 3.2 + (0.21)(6.4) = 4.544\text{ rad/s}$. For displacement, $\Delta\theta = \omega_i t + 1/2\alpha t^2 = (3.2)(6.4) + 0.5(0.21)(6.4)^2 = 20.48 + 4.30 = 24.78\text{ rad}$.
A 60. cm diameter wheel rotates through 50. rad. What distance will it move?
How many times will the wheel rotate?
A saw blade is spinning at 2000. rpm. (rpm = revolutions/min) What is its angular velocity?
Through how many radians will it spin in 2.0 seconds?
How fast are the blade tips moving if the blade has a 30. cm diameter?
A rotating satellite has the same angular velocity as the Earth. If the satellite is $5.0 \times 10^7$ m from the center of the earth, what is its tangential velocity?
4. Find the moment of inertia of a 200.-gram saw blade that is 30. cm in diameter.
5. What angular acceleration will result if the blade in #4 is acted upon by a 10. Nm torque? How long will it take to spin the blade up to its operating speed of 3000 rad/sec?
6. Pitcher problem: arm rotates stick mass + ball point mass. Torque required?
7. Boy pulls string on flywheel. Torque and angular acceleration?
TEACHER CER KEY (AP SCORE: 5/5)
Review Q1/Q2: Distance $s = 15\text{ m}$ and Rotations $= 7.96\text{ rev}$.
Reasoning: Radius $r = 30\text{ cm} = 0.30\text{ m}$. Linear distance $s = r\theta = (0.30)(50) = 15.0\text{ m}$. Rotations is the angular displacement divided by the displacement of one revolution: $50\text{ rad} / (2\pi\text{ rad/rev}) = 7.957\text{ rev}$.
Review Q6: $v_{sat} = 3636\text{ m/s}$.
Reasoning: Earth's angular velocity is one rotation per 24 hours: $\omega_E = (2\pi\text{ rad}) / (86400\text{ s}) = 7.27 \times 10^{-5}\text{ rad/s}$. Since the satellite has the same $\omega$, its tangential velocity is $v = r\omega = (5.0 \times 10^7\text{ m})(7.27 \times 10^{-5}\text{ rad/s}) = 3635.9\text{ m/s}$.
$= \Delta d / \Delta t$