#$&*
Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
When finding the moment of inertia for the strap plus the 2 magnets, do you add the masses together before m r^2 or do you add the separate moments of inertia to equal the I of the strap plus magnets?
You could add the masses of all magnets whose distances are the same, before squaring the r. But you couldn't do this with different values of r, because the r^2 would differ.
Also, when measuring the rubber bands, how are we to judge the difference between suspending the 1 domino and x? I'm confused as to how that plays into this experiment or better yet, how to determine it. For example, my chain is 6 rubber bands long and is 36 cm long when suspending 1 domino. How do I use that information to determine the average force?
Subtract 36 cm from the stretched length of the chain. Then multiply by k, using the rule I gave you to find k.
Thanks!
You're more than welcome.
#$&*
Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Regarding my timing of the ball up and down the ramp, I attempted to use TIMER and the only thing I can figure out is maybe I thought the wrong numbers corresponded to the distances since there are so many numbers listed, as I didn't have a watch with seconds at that time (silly 21st century). If I get a chance tonight, I will try to redo it as I now have a watch with seconds available.
The right-hand column gives you the intervals between clicks. A watch, unless it's a stopwatch, would be of limited value given the precision required by this experiment.
Coefficient of Friction--Steel
#$&*
course Phy 201
11/5 1:30 pmOne lab activity:
Conservation of Energy on a Ramp
Use an incline with a very small slope.
Strike the steel ball so it coasts up, comes to rest, then accelerates back down the incline. Let the ball continue to roll off the incline and fall to the floor, and mark the position at which it hits the floor. (Be sure you also mark the straight-drop position).
Also note the position on the incline at which the ball comes to rest before accelerating back down.
At the same time, using the TIMER program, record the time interval from the end of the 'strike' back to the end of the incline.
Repeat for at least two good trials.
Report your data:
****
5.984375 s, 6.84375s
#$&*
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:
****
Released at 6.5 cm on the ramp==1.171875 s, 6 cm from free drop
Released at 9 cm on the ramp==1.540625 s, 10.5 cm from free drop
This would indicate that after being started, the ball in the first trial required about 6 seconds to reach its stopping point, which was only 6.5 cm up the ramp, and return to its original position. That is possible. However the ball would have take about as long to come down as it did to travel up (the time down would in fact exceed the time up). That would be roughly 3 seconds up, and 3 seconds back. It's difficult to imagine that the ball would then, when released from the previous stopping point 6.5 cm up the ramp, accelerate down the ramp in about 1/3 the time it required after its previous stopping point.
#$&*
From your data, infer how long it took the ball to go up the ramp for each trial.
Calculate the ball's acceleration up the ramp, and its acceleration down the ramp.
Report your results, and indicate how they were obtained:
****
Trial 1: took 4.81s to travel 6.5 cm up, 1.17 s to travel down the ramp
A_up=-.56cm/s^2 a_down=9.4 cm/s^2
Trial 2: took 5.3 s to travel 9 cm up, 1.54 s down
A_up=-.64 cm/s^2 a_down=7.53 cm/s^2
To find acceleration, I used the distance traveled and the time to solve for Vave. For the portions up the ramp, the acceleration is negative due to the fact that Vf=0. After finding V0 or Vf, I used the time again to solve for acceleration.
#$&*
From your two accelerations, you can infer the coefficient of rolling friction. What is your result?
****
Trial 1--4.42
Trial 2—3.495
****I know this isn’t correct, but I am not sure what to do with my results to figure it out.
rolling friction up is mu * m g, directed down the ramp, so net force when traveling up is -slope * m g - mu * m g
rolling friction down is mu * m g, directed up the ramp, so net force when traveling down is -slope * m g + mu * m g
acceleration is net force / mass, so accelerations are respectively
-slope * g - mu * g and
-slope * g + mu * g.
The difference between the accelerations is 2 mu * g.
#$&*
The large ball has diameter 2.5 cm, the small ball diameter 2.0 cm. The ball is made of steel with an approximate density of 8 grams / cm^2.
Find the KE the ball attained while rolling down the ramp with the PE it lost while rolling down. Compare the two:
****
Trial 1:
KE=1/2 m (v^2)
1/2 (16g) (5.5 cm/s)^2=242 g cm^2/s^2
PE=16 g*9.4cm/s^2*6.5 cm=978g cm^2/s^2
The difference is 736 g cm^2/s^2
Trial 2:
KE=1/2 (16 g) ( 5.8 cm/s)^2=269 g cm^2/s^2
PE= 16 g*7.53 cm/s^2*9 cm=1084 g cm^2/s^2
The difference is 815 g cm^2/s^2
#$&*
Calculate the energy lost to rolling friction. What do you get?
****
#$&*
Can you account for all PE that was lost as the ball rolled down the ramp? Give details.
****
#$&*
I have a question about your data, which differ in nature from what we would expect. See my note, and see if you can clarify (or verify that you did time the expected events).
See also my note on how the difference in the two accelerations leads you to the coefficient of friction. You can also see Class Notes for 101108.
#$&*
** **
** **
See my note. We'll try to talk aobut this one in lab tomorrow.