8.01SC | Fall 2016 | Undergraduate

# Classical Mechanics

## Week 3: Circular Motion

« Previous | Next »

### Week 3 Problem Set

« Previous | Next »

« Previous | Next »

Acceleration in Circular Motion: $$\vec{a} = r\frac{d^2\theta}{dt^2}\hat{\theta} + r(\frac{d\theta}{dt})^2(-\hat{r})$$

For circular motion, show that $$\frac{d \hat{\theta}}{dt} = -\frac{d \theta}{dt}\hat{r}$$

We know that $$\hat{\theta }$$ changes direction as we go around a circle, and so if a particle is undergoing circular motion, we expect $$\frac{d\hat{\theta }}{dt}$$ to be non-zero. To figure out exactly what it is, let us write $$\hat{\theta }$$ in terms of $$\hat{i}$$ and $$\hat{j}$$ for an arbitrary $$\theta$$.

\begin{align} \hat{\theta} = -\sin(\theta) \hat{i} + \cos(\theta) \hat{j} \end{align}

Since $$\hat{i}$$ and $$\hat{j}$$ are constant anywhere in space, they are not time-dependent, so now the derivative becomes much more clear. Using the chain rule, we have:

\begin{align} \frac{d\hat{\theta}}{dt} = -\cos(\theta) \frac{d\theta}{dt} \hat{i} - \sin(\theta) \frac{d\theta}{dt} \hat{j} = - \frac{d\theta}{dt} \left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) \end{align}

We know that $$\left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) = \hat{r}$$, so we can now simply write this derivative as:

\begin{align} \frac{d\hat{\theta}}{dt} = - \frac{d\theta}{dt} \hat{r} \end{align}

« Previous | Next »

« Previous | Next »

« Previous | Next »

« Previous | Next »

A particle is moving in a circle of radius $$\displaystyle r$$ with an angular acceleration given by $$\displaystyle \vec{\alpha }(t)=\dfrac {1}{r}(A-Bt)\hat{k}$$, or $$\displaystyle a_{\theta }=(A-Bt)\hat{\theta }$$ where $$\displaystyle A$$ and $$\displaystyle B$$ are positive constants. At time $$\displaystyle t_0=0$$, the particle is at an angle $$\displaystyle \theta (0)$$ measured with respect to the $$\displaystyle +x$$-axis, and the tangential component of its velocity is $$\displaystyle v_{\theta }(0)=v_0$$.

Find the particle’s angular position $$\displaystyle \theta (t)$$ as a function of time to obtain the arc length travelled by the particle during the time $$\displaystyle t$$ and defined as $$\displaystyle s(t)=r(\theta (t)-\theta (0))$$. Express your answer in terms of $$\displaystyle r$$, $$\displaystyle A$$, $$\displaystyle B$$, $$\displaystyle t$$, and $$\displaystyle v_0$$

« Previous | Next »

« Previous | Next »

Position: $$\vec{r}(t)=r\hat{r}$$

Angular Velocity: $$\vec{\omega}=\omega_z \hat{k}$$

Component of the Angular Velocity along the axis of rotation: $$\omega_z=\dfrac{d\theta}{dt}$$

Velocity: $$\vec{v}=r\dfrac{d\theta}{dt}\hat{\theta}$$

Angular Acceleration: $$\vec{\alpha} = \alpha_{z}\hat{k}$$

Component of the Angular Acceleration: $$\alpha_z=\dfrac{d\omega_z}{dt}=\dfrac{d^2\theta}{dt^2}$$

Acceleration: $$\vec{a}=a_r\hat{r}+a_{\theta}\hat{\theta}$$

$$a_r = -\frac{v^2}{r} = -r(\dfrac{d\theta}{dt})^2, a_{\theta} = r\dfrac{d^2\theta}{dt^2}= r\alpha_z$$

« Previous | Next »

« Previous | Next »

« Previous | Next »

« Previous | Next »

A car of mass $$\displaystyle m$$ is turning on a banked curve of angle $$\displaystyle \phi$$ with respect to the horizontal. The curve is icy and friction between the tires and the surface is negligible. The curve has a radius $$\displaystyle r$$. What is the speed $$\displaystyle v$$ at which the car can turn safely? Express your answer in terms of $$\displaystyle g$$, $$\displaystyle r$$, and $$\displaystyle \phi$$.

« Previous | Next »

« Previous | Next »

In this demo, a bucket full of ping pong balls is rotated vertically fast enough so that the balls stay in the bucket. Slowing down below a certain speed, or simply holding the bucket upside down will cause the balls to fall. This can be explained using Newton’s second law and the physics of circular motion.

« Previous | Next »

« Previous | Next »

« Previous | Next »

« Previous | Next »

### Chain Rule of Differentiation

Recall that when taking derivatives of a differentiable function $$f = f(\theta)$$ whose argument is also a differentiable function $$\theta=g(t)$$ then $$f=f(g(t))=h(t)$$ is a differentiable function of $$t$$ and

$$\frac{df}{dt}=\frac{df}{d\theta}\frac{d\theta}{dt}$$

Note that this is only in the case of circular motion, where $$r$$ is a constant in time. Otherwise, our derivative of $$f$$ would also have a $$\frac{df}{dr}\frac{dr}{dt}$$ term.

One way to measure an angle is in radians. A full circle has $$2\pi$$ radians. This week, we will use radians to measure the angles, so all angles will have units of radians, angular velocity will have units of radians/s, and angular acceleration will have units of radians/s$$^2$$. If we multiply these by a distance, such as $$r$$, the units will be m, m/s, or m/s$$^2$$.

External References

« Previous | Next »

« Previous | Next »

« Previous | Next »

« Previous | Next »

Uniform Circular Motion: $$\frac{d\theta}{dt}=constant$$

The tangential component of the acceleration is zero.

The magnitude of the radial acceleration can be written as $$|a_r|=\frac{v^2}{r}=r \omega^2=r(2 \pi f)^2=\frac{4 \pi^2 r}{T^2}$$.

« Previous | Next »

« Previous | Next »

« Previous | Next »

« Previous | Next »

Period, $$T$$, is defined as the amount of time it takes to go around once - the time to cover an angle of $$2\pi$$ radians.

Frequency, $$f$$, is defined as the rate of rotation, or the number of rotations in some unit of time.

Angular frequency, $$\omega$$, is the rotation rate measured in radians.

These three quantities are related by $$f=\frac{1}{T}=\frac{\omega}{2\pi}$$.

The speed at which an object goes around a circle can be related to these quantities through $$v=R \omega=\frac{2\pi R}{T}$$.

« Previous | Next »

« Previous | Next »

Problem Set 3 contains the following problems:

1. Bead on a Rotating Hoop
2. Banked Turn
3. Tetherball Breaking Off
4. Two Boxes Around a Shaft
5. Satellite
6. A Coin on a Rotating Disk

« Previous | Next »

« Previous | Next »

A person on a spherical asteroid of mass $$\displaystyle m_1$$ and radius $$\displaystyle R$$, sees a small satellite of mass $$\displaystyle m_2$$ orbiting the asteroid in a circular orbit of period $$\displaystyle T$$.

Express you answer in terms of some or all of the following: $$\displaystyle m_1$$,$$\displaystyle m_2$$, $$\displaystyle \pi$$, $$\displaystyle T$$, and the universal gravitational constant $$\displaystyle G$$.

(Part a) What is the radius $$\displaystyle r_{\text {sat}}$$ of the satellite’s orbit?

### Worked Example - Orbital Circular Motion – Radius

(Part b) What is the magnitude of the velocity of the satellite?

Express you answer in terms of some or all of the following: $$\displaystyle m_1$$, $$\displaystyle m_2$$, $$\displaystyle \pi$$, $$\displaystyle T$$, and the universal gravitational constant $$\displaystyle G$$.

### Orbital Circular Motion - Velocity

(Part c) If the asteroid rotates about its axis with a period $$\displaystyle T_{\text {a}}$$, at what radius must the satellite orbit the asteroid so that the satellite appears stationary to the person on the asteroid?

Express you answer in terms of some or all of the following: $$\displaystyle m_1$$, $$\displaystyle m_2$$, $$\displaystyle \pi$$, $$\displaystyle T_ a$$, and $$\displaystyle G$$.

### Orbital Circular Motion - Period

« Previous | Next »

« Previous | Next »

« Previous | Next »

Fall 2016
Lecture Videos
Problem Sets
Online Textbook