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course PHY241
Conservation of Energy on a Ramp
Use an incline with a very small slope.
Strike the steel ball so it coasts up, comes to rest, then accelerates back down the incline. Let the ball continue to roll off the incline and fall to the floor, and mark the position at which it hits the floor. (Be sure you also mark the straight-drop position).
Also note the position on the incline at which the ball comes to rest before accelerating back down.
At the same time, using the TIMER program, record the time interval from the end of the 'strike' back to the end of the incline.
Repeat for at least two good trials.
Report your data:
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8cm, 2.438, 0.7cm
12cm, 3.109, 1.4cm
18cm, 3.875, 2cm
The first number is where the ball came to a rest on the incline (ruler started at 0cm at END of incline) and the second number is the time interval from when I nudged it upwards to the end of ramp. The third set of numbers is the positions it landed on the paper (measured from the ‘straight dropped’ position point.) I set up my incline by raising it with 2 nickels (coins) and lined the ruler to the side of it.
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From each of the stopping points on the ramp, as you previously observed them, release the ball from rest and time it down the incline. Record the positions at which it strikes the floor.
Report your data:
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8cm, 1.203, 2.5cm
12cm, 1.656, 2.8cm
18cm, 2.125, 3.4cm
The first number is the starting position on the incline (where it came to rest on previous question), the second number is the time interval from starting position to end of incline, and the third number is the position it landed on the paper, measuring from the ‘straight drop position point.’
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From your data, infer how long it took the ball to go up the ramp for each trial.
Calculate the ball's acceleration up the ramp, and its acceleration down the ramp.
Report your results, and indicate how they were obtained:
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up incline: 1.235(8cm), 1.453(12cm), 1.75(18cm)
accel up: (10.50cm/s^2), (11.37cm/s^2), (11.76cm/s^2)
accel down: (11.05cm/s^2), (8.75cm/s^2), (7.97cm/s^2)
To get how long it took for the ball to go up the incline, I just subtracted the down time from the total time it took to go up to the floor. I got the accelerations up by us [‘ds=1/2a’dt^2. I got the accelerations down by doing the same equation and except I used the time it took for the ball to go straight down.
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From your two accelerations, you can infer the coefficient of rolling friction. What is your result?
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1st trial (12cm): mu=0.0013
2nd trial (18cm): mu=0.002
avg mu=0.00165
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The large ball has diameter 2.5 cm, the small ball diameter 2.0 cm. The ball is made of steel with an approximate density of 8 grams / cm^2.
Find the KE the ball attained while rolling down the ramp with the PE it lost while rolling down. Compare the two:
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Sphere volume: (pi*diameter^3)/6, (pi*2^3)/6=4.189
Mass=d/v
Mass=(8g/cm^2)/(4.189)=33.51grams
Vf=a’dt
At 12cm: vf=(8.75cm/s^2)*(1.656sec)=14.5cm/s
At 18cm: vf=(7.97cm/s^2)*(2.125sec)=16.94cm/s
KE=1/2mv^2
At 12cm: KE=0.5(33.51g)(14.5^2)=3522.95 J
At 18cm: KE=0.5(33.51g)(16.94^2)=4808.4 J
‘dPE should equal -‘dKE. The PE lost by the ball is equal to the KE that’s gained.
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Calculate the energy lost to rolling friction. What do you get?
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Can you account for all PE that was lost as the ball rolled down the ramp? Give details.
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Looks good overall.
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