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PHYSICS, M5 EQ-Bank 25

In the 1840s, French physicist, Hippolyte Fizeau performed an experiment to measure the speed of light. He shone an intense light source at a mirror 8 km away and broke up the light beam with a rotating cogwheel. He adjusted the speed of rotation of the wheel until the reflected light beam could no longer be seen returning through the gaps in the cogwheel.

The diagram shows a similar experiment. The cogwheel has 50 teeth and 50 gaps of the same width.
 

Explain why specific speeds of rotation of the cogwheel will completely block the returning light. Support your answer with calculations.   (5 marks)

Show Answers Only
  • Light travelling through a gap in the cogwheel can be completely blocked by a tooth. This occurs if a tooth moves exactly the width of a gap in the time it takes for light to travel to the mirror and back.
  • Calculating the time taken:
  •    `t=(s)/(v)=(2xx8000)/(3.00 xx10^(8))=5.33 xx10^(-5)  text{s}`
  • To completely block the light, the tooth will have moved into the path of a gap in this time. Since there are 50 teeth and 50 gaps, the wheel will have undergone one hundredth of a rotation in this time.
  •    `omega=(Delta theta)/(t)=((2pi)/(100))/(5.33 xx10^(-5))=1180  text{rad s}^(-1)`
  • Additionally, the light will also be completely blocked if the cogwheel is spun at 3, 5, 7, or any odd multiple of this speed. In these cases, the wheel turns an odd number of gap-tooth intervals in the time it takes light to return.
  • For example, at three times this speed, the wheel rotates three hundredths of a full turn during the light’s round trip, so the returning light meets the second tooth instead of a gap and is blocked.
Show Worked Solution
  • Light travelling through a gap in the cogwheel can be completely blocked by a tooth. This occurs if a tooth moves exactly the width of a gap in the time it takes for light to travel to the mirror and back.
  • Calculating the time taken:
  •    `t=(s)/(v)=(2xx8000)/(3.00 xx10^(8))=5.33 xx10^(-5)  text{s}`
  • To completely block the light, the tooth will have moved into the path of a gap in this time. Since there are 50 teeth and 50 gaps, the wheel will have undergone one hundredth of a rotation in this time.
  •    `omega=(Delta theta)/(t)=((2pi)/(100))/(5.33 xx10^(-5))=1180  text{rad s}^(-1)`
  • Additionally, the light will also be completely blocked if the cogwheel is spun at 3, 5, 7, or any odd multiple of this speed. In these cases, the wheel turns an odd number of gap-tooth intervals in the time it takes light to return.
  • For example, at three times this speed, the wheel rotates three hundredths of a full turn during the light’s round trip, so the returning light meets the second tooth instead of a gap and is blocked.

PHYSICS, M5 EQ-Bank 27

A toy car was placed facing outwards on a rotating turntable. The car was held in place by a force sensor connected to the centre of the turntable. The centre of mass of the car was 0.25 metres from the centre of the turntable. The reading from the force sensor was recorded at varying speeds of rotation. A stopwatch was used to time the rotation of the turntable. The linear velocity was calculated from the period of rotation. The graph shows the force on the car versus the square of the linear velocity of the car.
 


 

  1. Use the graph to determine the mass of the car.   (3 marks)
  1. Identify possible errors in the data and outline how to reduce their effects on the estimation of the mass of the car.   (4 marks)
Show Answers Only

a.   0.034 kg

b.   Errors and reduction strategies:

  • If the sensor does not produce accurate readings then systemic errors will result. 
  • Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.
  • If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.
  • A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.
Show Worked Solution

a.   Graph passes through `(2,0)` and `(25, 3.1)`:

`text{Gradient}` `=(3.1-0)/(25-2)=0.135`  
`text{Gradient}` `=(F)/(v^2)=0.135`  

 
Since the car is undergoing uniform circular motion:

`F_(c)` `=(mv^2)/(r)`  
`(F_(c))/(v^2)` `=(m)/(r)`  
`0.135` `=(m)/(r)`  
`:. m` `=0.135 xx 0.25=0.03375=0.034\ text{kg}`  

 

b.   Errors and reduction strategies:

  • If the sensor does not produce accurate readings then systemic errors will result. 
  • Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.
  • If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.
  • A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.

PHYSICS, M5 2021 HSC 22

A horizontal disc rotates at a constant rate as shown. Two points on the disc, `X` and `Y`, are labelled. `X` is twice as far away from the centre of the disc as `Y`.
 

Compare the angular and instantaneous velocities of `X` with those of `Y`.   (3 marks)

Show Answers Only
  • The angular velocities `(omega)` of  `X` and `Y` are equal. The direction of instantaneous velocities for `X` and `Y` are the same.
  • Since `v=r omega` the magnitude of the instantaneous velocity of `X` is twice that of `Y`.
Show Worked Solution
  • The angular velocities `(omega)` of  `X` and `Y` are equal. The direction of instantaneous velocities for `X` and `Y` are the same.
  • Since `v=r omega` the magnitude of the instantaneous velocity of `X` is twice that of `Y`.