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Learning Introductory Physics with Activities

Paul J. Emigh, Rebecka Tumblin, Kathryn Hadley, Danielle Skinner

Section 11.11 Practice - Rotational Motion

Subsection Explanation Practice

Explanation 11.11.1. Acceleration of a Pendulum.

Suppose a pendulum is pulled to one side and released at \(t_1\text{.}\) At \(t_2\text{,}\) the pendulum has swung halfway back to a vertical position. At \(t_3\text{,}\) the pendulum has swung all the way back to a vertical position. Rank the three instants in time by the magnitude of the centripetal acceleration, from greatest to least.

Subsection A*R*C*S Practice

A*R*C*S 11.11.2. Swinging a Bucket.

You swing a bucket of water with total mass m in a vertical circle with a radius given by the length of your arm, \(L\text{.}\) Determine the slowest speed at which you can swing the bucket without water falling on you. For that speed, determine the tension in your arm \(T\) at the bottom of the swing.

A*R*C*S 11.11.3. Conical Pendulum.

A conical pendulum is formed by attaching a ball of mass \(m\) to a string of length \(L\text{,}\) then allowing the ball to move in a horizontal circle of radius \(r\text{.}\)
Part A. Find an expression for the tension \(T\) in the string.
Part B. Find an expression for the ball’s angular speed \(\omega\text{.}\)

A*R*C*S 11.11.4. Car Around a Curve.

A car starts from rest on a curve with a radius of \(120 \mathrm{~m}\) and tangential acceleration of \(1.5 \mathrm{~m/s^2}\text{.}\) Through what angle will the car have traveled when the magnitude of its total acceleration is \(3 \mathrm{~m/s^2}\text{?}\)
Tip.
For your representation (part 1c), sketch the different components of the acceleration vector for the car.
Solution.
1. Analyze and Represent
  1. List quantities.
    \begin{equation*} R = 120 \mathrm{~m} \end{equation*}
    \begin{equation*} a_t = 1.5 \mathrm{~m/s^2} \end{equation*}
    \begin{equation*} |a| = 3 \mathrm{~m/s^2} \end{equation*}
    \begin{equation*} \omega_i = 0 \end{equation*}
    \begin{equation*} \omega_f = ? \end{equation*}
    \begin{equation*} \Delta\theta = ? \end{equation*}
  2. Identify assumptions. The motion of the car is circular and its angular acceleration is constant.
  3. Represent the situation physically. The figure below shows the tangential and centripetal acceleration for the final position of the car, when the magnitude of the acceleration is \(3 \mathrm{~m/s^2}\text{.}\) At this instant, the centripetal acceleration should be larger than the tangential acceleration in order for the two components to add together properly.
Figure 11.11.1. Acceleration components for the final position of the car.
2. Calculate
  1. Represent principles symbolically.
    \begin{equation*} a_c = R\omega^2 \end{equation*}
    \begin{equation*} a_t = R\alpha \end{equation*}
    Since the angular acceleration is constant:
    \begin{equation*} \omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta \end{equation*}
    Lastly, the magnitude of the acceleration vector is:
    \begin{equation*} |a| = \sqrt{a_t^2 + a_c^2} \end{equation*}
  2. Solve unknown(s) symbolically. The tangential acceleration is known and constant, so the centripetal acceleration at the end is the thing you want to know.
    \begin{equation*} a^2 = a_t^2 + a_c^2 = a_t^2 + R^2\omega_f^4 \end{equation*}
    \begin{equation*} a^2 = a_t^2 + R^2(2\alpha\Delta\theta)^2 \end{equation*}
    \begin{equation*} a^2 = a_t^2 + 4R^2\alpha^2(\Delta\theta)^2 \end{equation*}
    \begin{equation*} (\Delta\theta)^2 = \frac{a^2 - a_t^2}{4R^2\alpha^2} \end{equation*}
    \begin{equation*} \Delta\theta = \sqrt{\frac{a^2 - a_t^2}{4a_t^2}} \end{equation*}
  3. \begin{equation*} \Delta\theta = 0.866 \mathrm{~rad} \end{equation*}
3. Sensemake
  1. Are the units correct?
    \begin{equation*} \Delta\theta = \sqrt{\frac{a^2 - a_t^2}{4a_t^2}} \end{equation*}
    \begin{equation*} \mathrm{rad} = \sqrt{\frac{\mathrm{m^2/s^4}}{m^2/s^4}} = \mathrm{unitless} \end{equation*}
    Radians are unitless, so this checks out!
  2. Is your numerical answer reasonable? This is just under one radian, so just under one sixth of a full circle. At first glance, this might seem small, but the car is moving around a pretty big circle, and will in fact have traveled just over 100 m by the time it finishes this acceleration, which seems like a reasonable distance for a car to move at everyday speeds and accelerations.
  3. Does your symbolic answer make physical sense?
    There are a lot of interesting possibilities for making sense of this symbolic answer, but the one that stands out is the numerator, which is a difference between the squares of the accelerations. In particular, the special case where \(a = a_t\) makes the numerator vanish, which means \(\Delta\theta = 0\text{.}\) This makes sense from a physical standpoint because \(a = a_t\) corresponds to the acceleration being entirely tangential, which is only true when the velocity is zero, before the car starts moving, when it will not have traveled any (angular) distance at all!

Subsection Numerical Practice

Calculation 11.11.5. Crankshaft.

The crankshaft (\(r = 3 \mathrm{~cm}\)) in a race car accelerates uniformly from rest to \(3000 \mathrm{~rpm}\) in \(2.0 \mathrm{~s}\text{.}\) How many revolutions does it make?
  1. 10 revolutions
  2. 25 revolutions
  3. 42 revolutions
  4. 50 revolutions
  5. 63 revolutions
  6. 101 revolutions
Answer.
D.

Calculation 11.11.6. Angular motion.

A 60 cm diameter wheel accelerates from rest at a rate of \(7 \mathrm{~rad/s^2}\text{.}\)
  1. What is the tangential acceleration of a point on the edge of the wheel?
  2. How long does it take for the wheel to turn through 14 rotations?
  3. After the wheel has undergone 14 rotations, what is the radial component of the acceleration on the edge the wheel?
Answer 1.
\(2.1 \mathrm{~m/s^2}\)
Answer 2.
\(5.01 \mathrm{~s}\)
Answer 3.
\(369 \mathrm{~m/s^2}\)

Calculation 11.11.7. Army the Armadillo.

The Brazilian three-banded armadillo has a hard outer exterior, and when it senses danger, it rolls into a ball for protection. Suppose that Army the armadillo rolls into a ball (with a \(5 \mathrm{~cm}\) radius) while on the slope of a hill. As a result, they begin to roll down the hill; and after \(15 \mathrm{~s}\text{,}\) they have undergone \(45.8\) revolutions. What is Army’s angular speed at this time?
  1. \(\displaystyle 12.7 \mathrm{~rad/s}\)
  2. \(\displaystyle 15.5 \mathrm{~rad/s}\)
  3. \(\displaystyle 22.0 \mathrm{~rad/s}\)
  4. \(\displaystyle 29.1 \mathrm{~rad/s}\)
  5. \(\displaystyle 38.4 \mathrm{~rad/s}\)
How fast is Army traveling at this point?
  1. \(\displaystyle 1.92 \mathrm{~m/s}\)
  2. \(\displaystyle 2.23 \mathrm{~m/s}\)
  3. \(\displaystyle 3.49 \mathrm{~m/s}\)
  4. \(\displaystyle 4.24 \mathrm{~m/s}\)
  5. \(\displaystyle 5.56 \mathrm{~m/s}\)
Answer 1.
E.
Answer 2.
A.

References References

[1]
Numerical practice activities provided by BoxSand: https://boxsand.physics.oregonstate.edu/welcome.
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