I wanted to see if some of my students could work out the inductance of a coil i wound around an iron core using any of the equipment they wanted in the lab. Before that I had shown them the non-ideal nature of the coil by connecting it in series with a DC ammeter and battery and the ammeter showed a few amps.

Kate gave it a go by selecting a variable capacitor with accurate scale of 100-300 microFarad. She connected this in series with the coil to a 12V/50Hz AC supply. She monitored the voltage across the coil and the capacitor, carefully adjusting the latter until the two voltages are exactly the same. The capacitance read 219 microFarad. Using the equation

f = 1/2pi(LC)*1/2

she got a value of 46.3 mH for the inductance

Sharon did something different. She chose a 6.00 ohm resistor and connects it in series with an AC ammeter and the 12V/50Hz AC supply. She records a current of 0.657 A and the voltage across the resistor is 3.94 V and across the coil it is 9.53 V

Have you any idea how she used this figures to get the correct inductance of the coil?

Have a go and let me know.

Be critical of Kate's expreriment

From our upstairs Physics lab, we look out over a lawn bowls club and , now that spring has just about arrived, the lawn has been beautifully mown and the old folk who play have dusted off their white clothes and begun to play.

This led me to use the lawn bowl as a teaching tool for the girls/boys in y13.

Aside view of a lawn bowl is circular while the front on view shows their is more mass on one side. When a bowl stops it generally falls over and I asked them to explain this.

Kate :- " Where the bowl touches the ground is where the reaction force acts upwards. The gravity force acts through the centre of mass which is to the heavy side of the bowl, so there is a torque acting which is not balanced so it falls over."

( not only the best looking student with the shortest skirt, Kate is the brainiest).

We took some video of a bowl falling over and we were able to work out that it had a rotational velocity of 2.8 rad/s and Inertia of .00216 kg/m2.

Using Kinetic energy (rot.) = 1/2Iw2 , Kinetic energy (lin.) = 1/2 mv2 ( v = rw )

we got values of 8.5 mJ for rotational E and 21mJ for linear E and that the centre of mass of the bowl drops by about 2mm

When I asked them to explain why the bowl has a curved path when bowled, again Kate was to the fore. Distractingly shifting in her seat she declared; " the bowl wants to tip over in one directon so the friction force from the grass acts in the opposite way. When its travelling upright, this friction force is the only force acting on it, so shifts it to one side'.

When the bell goes for the end of class and the uncoil themselves from their seats, I sigh.

Next time I might get them to see that a bowl rocking on its side would do so with Simple Harmonic Motion.

Kapai

A cylindrical satellite has a camera that it can rotate around its centre on a string that can be drawn in or out. If the radius of the satellite is R, the longest length of the string is rl rotating with velocity vl, then at any string length r with velocity v we can draw a relationship. If we regard the mass of the camera as being negligable when compared with the satellite, then angular momentum will be conserved

L = mvr

so, mvr = mvlrl so v = vlrl/r

there is a slot in the centre of the satellite so that the camera can get within the radius of the satellite as the string is drawn in. The tension force on the string is the centripetal force

F = mv^2/ r substituting the previous value for v gives us F = mvl^2rl^2/r3

this cube inverse relationship means the tension force on the string will increase rapidly as the string gets shorter- needs to be strong.

The amount of work to be done to bring the camera into the satellite will be the difference between the kinetic energy at length rl and that at length R

i.e. Change in E = 1/2 mv62 - 1/2 mvl^2

when I substitute in this I get change in E = mvl^2(rl^2-R^2)/R^2

London Eye - the giant rotating wheel with 32 capsules makes for a good physics study as it involves all sorts of questions in mechanics. The capsules have a mass of 10,000 kg and are 68 m from the centre. When moving they travel at a constant speed of 0.26 m/s. Consequently the centripetal force can be calculated from F = mv*2/r - (10,000)(.26)*2/68 = 9.9 N. ( not very different from earths gravitational force).

The angular speed in rad/s can be shown from v=rw as .0038 and the time for one complete rotation ( 2 pi r ) is about 27 minutes.

When it starts up it needs an average net torque of 46 million newton metres to accelerate it up to speed - i.e. an average angular acceleration of .0017 rad/s/s

In that start up time it rotates through 1/4 of a degree.

As the wheel increases its speed the frictional forces acting against the motion increases causing the unbalanced torque and the angular acceleration to decrease. When the torques reach balance there will be no further acceleration and the wheel now travels at constant speed. So there is no change in kinetic energy and the motor is simply matching the friction.

When a capsule is at the top it can sway a maximum of 0.15 m ie in SHm of 1.8 rad/s

If I was sitting in a capsule with this amount of sway would be moving sideways at about 2.6 m/s.at the start of the swing. Dampers are used to restrict the sway - otherwise the sway could be as much as 3 m in a strong wind.