2002 Autumn Basic Physics
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1. [33 points] Table on Springs

A. [25 points] A uniform table of total mass M rests on two springs, each with spring constant k. Only vertical motion of the table is allowed. The table is initially at rest. A soft lump of clay, of mass m, falls under the influence of gravity and hits the center of the table with velocity $v_c$ and sticks. The table starts oscillating. What is the table’s amplitude of oscillation? Ignore friction or other damping forces.

Greg’s Solution:

First, we observe that since the springs are in parallel they act as if they were a single spring with spring constant $2k$.

The initial displacement of the table from equilibrium is given from the equation

\begin{align} 2kx = -Mg \quad \Rightarrow \quad x = \frac{-Mg}{2k} \end{align}

After collision, the table has velocity

\begin{align} v = \frac{m}{m+M}v_c \end{align}

Thus, the total energy of the system is

\begin{align} \frac{1}{2}(2k)x^2 + \frac{1}{2}(m+M)v^2 = \frac{M^2g^2}{2k} + \frac{v_c^2}{2(m+M)} \end{align}

Since there is no dissipation, this is the same energy the system will have at its maximum amplitude, so that

\begin{align} \frac{1}{2}(2k)x_{max}^2 = \frac{M^2g^2}{2k} + \frac{v_c^2}{2(m+M)} \quad \Rightarrow\quad x_{max} = \sqrt{\frac{M^2g^2}{2k^2} + \frac{v_c^2}{2k(m+M)}} \end{align}

B. [8 points] Some time later the lump of clay is removed from the surface of the table. The clay is removed just as the table reaches its maximum height (the springs are maximally extended). Assume the clay is removed without any impulse to the table. Describe, qualitatively, the subsequent motion of the table.

The motion of the table speeds up, since the angular frequency of a harmonically oscillating spring is $\sim \sqrt{\frac{k}{\text{mass}}}$, and after the clay is removed the mass has decreased and therefore the frequency should increase.

The amplitude of the motion, however, decreases, because the system does not have as much energy as it used to and therefore the maximum displacement from equilibrium cannot be as great. Furthermore, the “center” of the oscillations will be a little higher, since the equilibrium position for the table is higher than it had been with the clay on top of it.

2. [33 points total] Cyclotron

A cyclotron, shown below, is made up of two hollow D-shaped electrodes (called Dees because they look like the capital letter D). A charged particle is initially placed in the center $+q$, and an oscillating potential with maximum magnitude $\Delta V$ (volts) is applied across the Dees. The Dees are immersed in a uniform magnetic field that is perpendicular to their plane (in or out of the page in the diagram) , and has a magnitude B. Assume that $\omega$, the angular frequency of $\Delta V$, is correctl tuned so that the magnitud of the voltage difference while the particle is between the two Dees is always $\pm\Delta V$. The two Dees are separated by 1 cm.

A. [5 points] Specify the direction of the magnetic field for the cyclotron to work as shown (with the particle exiting in the direction shown at the top of the Dees).

Greg’s Solution:

The magnetic field must be such that when the particle is travelling to the right, it is deflected down, and when it is travelling left, it is deflected up, since this will give it the desired clockwise motion. Using the right-hand rule, we see that this requires the magnetic field to be out of the page.

B. [14 points] Derive the cyclotron radius as a function of the particles velocity, v, and its mass m, and the external magnetic field B.

Greg’s Solution:

When the particle is travelling with velocity v, the force due to the magnetic field will be

\begin{equation} F_B = qvB \end{equation}

When the particle exits, we want it to be travelling in a uniform circle with radius equal to that of the cyclotron, so we want this centripetal force to equal the magnetic force,

\begin{align} \frac{mv^2}{R} = qvB \quad \Rightarrow \quad R=\frac{qB}{mv} \end{align}

C. [14 points] What is the cyclotron radius as a function of the number of times the particle crosses the gap between the Des? The particle starts at the center of the gap for the first crossing.

Greg’s Solution:

The kinetic energy of the particle increases by $q\Delta V$ each time it passes through the gap, save for the first time when it only receives $q\Delta V/2$ since it is located at the center. Thus, the kinetic energy after $N$ passes will be

\begin{align} K = \frac{1}{2}mv^2 = \left(N+\frac{1}{2}\right) q \Delta V \quad \Rightarrow \quad v = \sqrt{\frac{2}{m}\left(N+\frac{1}{2}\right) q \Delta V }. \end{align}

Plugging this into [[eqref cyclorad]], we obtain

\begin{align} R=\frac{qB}{m\sqrt{\frac{2}{m}\left(N+\frac{1}{2}\right) q \Delta V }} \end{align}

3. [[33 points total]] Optics

A. The Keplerian telescope (also known as refracting telescope) is used to study the planet Mars from Earth. The eye piece focal length is $f_e$, and the objective lens focal length is $f_o$.

i. [10 points] What is the proper distance between the objective and eye-piece lenses so that objects very far away (Mars) can be correctly imaged by a normal eye looking through the lens?

Greg’s Solution:

Since the image is very far away, the rays are nearly parallel. Thus, the image rays will be focused so that they converge at a point that is the focal distance away from the objective lens. We want these rays to be made parallel again after they pass through the eyepiece lens, so we need the eyepiece lens to be its focal length away from the point where the rays converged. Thus, the total distance between the two lenses should be

\begin{equation} L = f_0 + f_e \end{equation}

ii. [15 points] Derive the angular magnification for this telescope in terms of the two focal lengths.

B. [8 points] On a hot, sunny day you stand on a road and some distance away you observe a motorcycle approaching you. If you look carefully, it appears there is a reflection of the motorcycle on the road. This is one common example of a mirage. Explain this phenomenon qualitatively.

4. [33 points total]] Bowling Ball and Friction

A mechanical piston is used to shoot a bowling ball down a special alley lane. The bowling ball has radius R and mass M. The first meter of the alley is frictionless, and beyond that the coefficient of dynamic friction between the ball and the lane’s hardwood floor is $\mu$. The piston hits the bowling ball imparting instantaneous impulse. The moment of inertia for a solid sphere of radius $R$ mass $M$ is $I_{\text{sphere}}=\frac{2}{5}MR^2$.

A. [25 points] If the piston hits the bowling ball at the equator, at what time, t, will the ball no longer slip on the lane? Assume $t=0$ when the ball reaches the boundary between frictionless and non-frictionless surfaces of the alley.

Greg’s Solution:

Assume that the ball is imparted with some initial velocity $v_0$. When it hits the friction surface, it experiences a force of magnitude $\mu g M$ so that

\begin{align} M \dot v = -\mu g M quad \Rightarrow \quad v_0 - \mu g t \end{align}

Likewise, it receives a torque $\mu g M R$ so that

\begin{align} I \dot \omega = \frac{2}{5}MR^2 \dot \omega = \mu g M R \quad \Rightarrow \quad \omega = \frac{5 \mu g t}{2R}. \end{align}

The ball will stop slipping when

\begin{align} v = \omega R \quad \Rightarrow \quad t = \frac{7v_0}{2 \mu g} \end{align}

B. [8 points] How far above or below the equator ($h$) of the bowling ball should the piston hit so that the bowling ball rolls without slipping initially?

Greg’s Solution:

Assume the stick imparts some linear impulse $I$ creating linear momentum $Mv$; in so doing, it also imparts an angular impulse $Ih$ creating angular momentum $I\omega$. Thus, we obtain the relation,

\begin{align} M v = \frac{2 M R^2 \omega}{5h} \end{align}

If we want the ball to roll without slipping then we need to have $v=R\omega$, which the above relation implies that

\begin{align} h = \frac{2}{5}R \end{align}
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