Physics Experiment Report 2001

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Physics Experiment Report 2001
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Aim of the experiment
?
Which one will reach the bottom first?
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Aim of the experiment
Investigating

• relationship of the moment of inertia and the
  translation velocity of a ring and disk
• conservation of energy

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Procedure

• Set up the runway as shown on the table
• Attach the photogate at the lower end of the
  runway
• Suppose the height of the runway be h.
• Release the ring from the upper end
• measure the velocity when the ring reach the
  floor

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Data and Results

• md mr Rd Rr
• Length of the disk and the ring that passes
  the sensor
• L 8.9410-2 m

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For Disk
Data and Results

• Time/t (s) Speed L / t (ms-1)
• 0.0946 0.94548
• 0.0955 0.93657
• 0.1022 0.87689
• 0.0974 0.91838
• Average speed 0.91931

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Data and Results
For Ring

• t1 t2 t3 tt1-t2t3
• 1.4713 1.3855 0.0107 0.1025
• 1.6685 1.5740 0.0108 0.1053
• 2.0759 1.9822 0.0108 0.1045

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Time/t (s) Speed l / t (ms-1)0.1025 0.87261
0.1053 0.84941 0.1045 0.85591 Average
speed 0.85931
Data and Results
For Ring
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Data and Results
Observation

• So it can be seen that the final speed of the
  disk is larger than that of the ring.
• i.e. Vd gt Vr .(1)

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Conclusion
First investigation

• Do you know which one with higher value of moment
  of inertia ?
• The disk or the ring?

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Conclusion
First investigation

• Formula to calculate the moment of inertia of the
  disk is
• I ½ ( m R2 )
• Now md 0.2 kg , Rd 0.06 m,
• Id ½ 0.2 ( 0.06 ) 2 3.6 10-4

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Conclusion
First investigation

• Formula to calculate the moment of inertia of the
  ring is
• I ½ m ( Ro2 Ri2 )
• Now mr 0.2 kg , Ro 0.06 m, Ri 0.052 m
• Ir ½ 0.2 ( 0.062 0.0522 ) 6.304 10-4

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Conclusion
First investigation

• ? Id lt Ir ....(2)
• Vd gt Vr ....(1)
• From (1) and (2), we can see that

Smaller is the moment of inertia larger is the
final speed.
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Conclusion
Second investigation

• Total kinetic energy of the disk or the ring
• K.E. ½ mv2 ½ I w2
  ½ mv2 ½ I (V/r)2
• Loss in potential energy
• P.E. mgh

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Conclusion
Second investigation

• So we should prove the following equation with
  the measured data.

mgh ½ mv2 ½ I
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Conclusion
Second investigation

• Now lets discuss the disk first.
• md 0.2 kg h 6.5 cm 0.065 m R 0.06 m
• Id 3.6 10-4 Vd 0.91931 ms-1
• LHS mgh 0.2 10 0.065 0.13 J
• RHS ½ mv2 ½ I (V/R)2
• ½ 0.2 ( 0.91931 ) 2 ½ 3.6
  10-4 ( 0.91931/0.06 )2
  
• 0.13 J

• Energy is conserved ( for the disk ).

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Conclusion
Second investigation

• For the ring
• mr 0.2 kg h 6.5 cm 0.065 m R
  0.06 m
• Ir 6.304 10-4 Vr 0.85931ms-1

LHS mgh 0.2 10 0.065 0.13 J RHS ½
mv2 ½ I (V/R)2 0.13 J

• Energy is conserved ( for the ring).

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Conclusion
First Second investigation
mgh ½ mv2 ½ I (V/R)2
From this equation, we can see that smaller
moment of inertia ( I ) gives larger value of V.
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Error Resources / Experiment Precautions

• Air resistance acting on the disk/ring causes
  energy loss
• The disk/ring is not rolling in a straight line
• Human measuring error
• The disk/ring may be rolling down with slipping
  which will cause energy loss by the friction

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The End
Bye!

• Thats the end for our presentation

B u t
How can we adjust the ring or disk so the they
reach the bottom at the same time?