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Section 11.5 Noninertial Frame Bootcamp

Subsection 11.5.1 Accelerating Frame

Subsection 11.5.2 Rotating Frame

Subsection 11.5.3 Rotation of Earth

Subsection 11.5.4 Miscelaneous

A rocket of mass \(M\) is moving at constant velocity \(\vec v_0\) in zero gravity environment. It ejects fuel from the back at a steady rate of \((dm/dt = \alpha )\) at speed \(u\) with respect to the rocket. What is the increase in speed of the rocket when a mass \(m_f\) of fuel has been ejected? Do this problem in the frame of the rocket.



See solution.


In the rocket frame, the equation of motion will have a fictitious force on the rocket. The rocket in the question is moving in the outer space since we do not have any gravity on the rocket. Let the \(x\)-axis of the rocket frame be pointed towards the forward direction of the rocket as shown in the figure. In the rocket frame, the velocity of the rocket be zero and the velocity of the burnt fuel will be \(-u \hat u_x\text{.}\)

We look at the change in the momentum of the rocket plus fuel in an time interval \(\Delta t\text{.}\) Let the \(x\)-component of the average acceleration of the rocket be \(\Delta v_x/\Delta t\text{,}\) where \(\Delta v_x\) is the change in the \(x\)-component of the velocity.

The equation of motion of the rocket written as a change in momentum over an interval \(\Delta t\) will be

\begin{equation*} F_{x}^{ext} \Delta t - m \Delta v_x = -|\Delta m| u, \end{equation*}

where \(\Delta m\) is the change in the mass of the rocket due to the ejection of the fuel. Here \(\Delta m\lt 0\text{.}\) With external force zero this equation becomes.

\begin{equation*} m \Delta v_x = - u \Delta m. \end{equation*}

Figure 11.5.21.

Writing this as infinitesimal \(dv_x\) and \(dm\) allows us the integration needed to find the change over an interval.

\begin{equation*} \int_{v_0}^{v_x(t)} d v_x = - u \int_M^{m(t)}\frac{dm}{m}. \end{equation*}

This gives

\begin{equation*} v_x(t) - v_0 = u \ln\left( \frac{M}{m(t)}\right). \end{equation*}

Since mass is ejected at a constant rate \(\alpha\text{,}\) we have \(m(t) = M-\alpha t\text{.}\) This gives,

\begin{equation*} v_x(t) - v_0 = u \ln\left( \frac{M}{M-\alpha t}\right). \end{equation*}

Here \(\alpha t\) is the mass of the ejected fuel. The question asks about the change in velocity when the mass ejected is equal to \(m_f\text{.}\) Let \(t=T\) at this instant. Then we can write the change in the \(x\)-component of the velocity as

\begin{equation*} v_x(T) - v_0 = u \ln\left( \frac{M}{M-m_f}\right). \end{equation*}

Since other components of the velocity are zero, this result also gives the change in the speed of the rocket during this interval.

A stone is tied to a string and rotated in a vertical circle at constant speed \(v\) inside an accelerating train. The circle of rotation is perpendicular to the train's velocity as observed by a passenger in the train who is at rest with respect to the train. The train has a constant acceleration of magnitude \(A\) and the direction towards the East with respect to an observer outside the train. The train was at rest at \(t=0\text{.}\)

Find the tension in the string at four instances in the vertical circle of the motion of the stone: (a) when the stone is at the top, (b) when the stone is at the bottom, (c) when the stone is horizontal left, and (d) when the stone is horizontal right.



See solution.


(a) The equation of motion in the train frame is easier to work out. The figure shows the forces on the stone when it is at the top of the circle. At that instant the centripetal acceleration is pointed down and has the magnitude \(v^2/R\) and while there does not appear to be any force along the horizontal direction, there is a non-zero fictitious force of magnitude \(mA\) pointed towards the negative \(x\)-axis.

Therefore, the tension must also have an \(x\)-component to go along with the fictitious force. Let \(T_x\) and \(T_y\) be the components of the tension along the \(x\)- and \(y\)-axes respectively.

\begin{align*} \amp T_y - mg = - m \frac{v^2}{R}.\\ \amp T_x -mA = 0. \end{align*}
Figure 11.5.23.

These equations give the following for the components of the tension.

\begin{align*} \amp T_x = mA. \\ \amp T_y= mg - m \frac{v^2}{R}. \end{align*}

Here \(T_y\le 0\) since the tension force on the stone will act away from the stone, which will imply that it will have a negative \(y\)-component here. We can combine the two components and determine the magnitude and direction of the tension force. This shows that when the stone is rotating in the circle, the string will be going over a conical surface that is pointed forward. The observer outside the train will find the same value of the tension, which will explain the forward acceleration of the the stone with respect to that observer.

(b), (c) and (d) are left for the student. Try similar arguments.

You are inside a large enclosed container and everything in the entire container, including you, is rotating at a constant rate about some fixed axis, but you do not know the rate of rotation. To find the rate of rotation you place a penny at different places on a frictionless floor, and discover that the penny accelerates everywhere you place the penny. The directions of the acceleration of the penny at various place on the flat floor meet at a point \(X\text{.}\)

(a) If penny accelerates at \(30\ \text{m/s}^2\) at a distance of 10 meters from the point \(X\text{,}\) what is the rate of rotation of the entire container? (b) What is the acceleration of the penny when it is at a point \(20\) meters from the point \(X\text{?}\)



(a) \(\sqrt{3}\ \text{rad/s}\text{,}\) (b) \(60\ \text{m/s}^2\text{.}\)

Solution 1 (a)

(a) In the rotating frame the acceleration is provided by the fictitious force \(-m\vec \Omega\times(\vec \Omega\times \vec r)\text{.}\) Here \(\vec \Omega\) is perpendicular to \(\vec r\text{.}\) Therefore, the magnitude of the acceleration will be \(\Omega^2 r\text{.}\) By equating this to the given acceleration we obtain

\begin{equation*} \Omega^2 \times 10\ \text{m} = 30\ \text{m/s}^2\ \ \Longrightarrow \ \ \Omega = \sqrt{3}\ \text{rad/s}. \end{equation*}
Solution 2 (b)

(b) At the point that is \(20\text{ m}\) from the axis through the point \(X\) we will have the acceleration of magnitude

\begin{equation*} \Omega^2 \times 20\ \text{m} = 60\ \text{m/s}^2. \end{equation*}

A pendulum of length \(L\) and mass \(m\) in an accelerating car is found to be in equilibrium at an angle \(\theta_0\) with respect to the vertical instead of the straight down direction. (a) What is the acceleration of the accelerating frame with respect to an inertial frame? (b) When the pendulum is disturbed from the equilibrium angle by a small angle \(\theta\) about \(\theta_0\text{,}\) the pendulum oscillates? Determine the frequency of small oscillations in the accelerating frame.



(a) \(g\tan\theta_0\text{,}\) (b) \(\frac{1}{2\pi}\sqrt{\frac{g}{L}\cos\theta_0}\text{.}\)

Solution 1 (a)

(a) When a pendulum is in an inertial frame fixed to the ground, the pendulum points down when not swinging. However, when the pendulum is in a non-inertial frame of constant acceleration, it will make an angle \(\theta_0\) when not swinging. This angle \(\theta_0\) will be related to the acceleration of the frame which we can find by either working with respect to the inertial frame \(Oxy\) or with respect to the non-inertial frame \(O'xy'\)shown in figure below. I will show the calculation in the non-inertial frame and you should work out the calculation for the inertial frame.

Figure 11.5.26.

Working in the non-inertial frame. In this frame the net force on the bob will be zero, but we have to include the fictitious force \(-m\vec A\text{.}\)

\begin{align*} \amp x\text{-component: }\ \ -T\sin\theta_0 +m A = 0.\\ \amp y\text{-component: }\ \ T\cos\theta_0 - m g = 0. \end{align*}


\begin{equation*} A = g\tan\theta_0. \end{equation*}
Solution 2 (b)

(b) Let the direction of the bob from the suspension point at some instant \(t\) be \(\theta\text{.}\) We will take \(\theta>\theta_0\) for concreteness in the calculations. The equations of motion in the non-inertial frame will now be

\begin{align*} \amp x\text{-component: }\ \ -T\sin\theta +m A = m\frac{d^2x'}{dt^2}.\\ \amp y\text{-component: }\ \ T\cos\theta - m g = m\frac{d^2y'}{dt^2}. \end{align*}

Although these equations are OK to work with but a better set will be the radial and angular coordinates of the bob with the origin at the suspension point rather that where it is shown in the figure. With this choice the angular equation of motion is written more readily as torque equation.

\begin{equation} -mg L \sin\theta = mL^2 \frac{d^2 \theta}{dt^2}.\label{eq-ch11-pr6-1}\tag{11.5.1} \end{equation}

Since \(\theta=\theta_0\) is the equilibrium angle, we define an angle \(\psi\) with respect to the equilibrium angle as

\begin{equation*} \psi = \theta - \theta_0. \end{equation*}

This gives the following for the small angle approximation for \(\psi\text{.}\)

\begin{equation*} \sin\theta = \sin(\theta_0+\psi) = \sin\theta_0\cos\psi + \cos\theta_0\sin\psi \approx\sin\theta_0+ (\cos\theta_0)\psi, \end{equation*}

Making the substitution in Eq. (11.5.1) we obtain

\begin{align*} \frac{d^2 \psi}{dt^2} \amp = -\left(\frac{g}{L}\cos\theta_0 \right) \psi - \frac{g}{L}\sin\theta_0\\ \amp = -\left(\frac{g}{L}\cos\theta_0 \right)\left( \psi - \frac{A}{g}\right) \end{align*}

Let us define a displaced angle by

\begin{equation*} \beta = \psi - \frac{A}{g}. \end{equation*}

In terms of \(\beta\) the equation of motion becomes

\begin{equation*} \frac{d^2 \beta}{dt^2} = -\left(\frac{g}{L}\cos\theta_0 \right) \beta \equiv - \omega^2 \beta. \end{equation*}

Therefore, the frequency of small oscillations about the equilibrium will be

\begin{equation*} f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{g}{L}\cos\theta_0}. \end{equation*}