0
$\begingroup$

I did a lab where we did this experiment with four different solids. It's a conservation of energy lab, so we calculated the potential energy at the beginning of the incline ($mgh$) and the kinetic energy at the end, just before it hits the ground. ($1/2 mv^2 (1+f)$) Here are those calculations for a spherical marble with a mass of 4.9 grams: $$PE=mgh=(4.9 [g])(9.8 [m⁄(s^2]))(0.089[m])=4.27 [mJ]$$ Finding the total kinetic energy of the marble at the end of the descent: $$KE_{tot}=KE_{trans}+KE_{rot}$$ $$=1/2 mv^2 (1+f)$$ $$=1/2(4.9 [g](0.38 [m]/0.419 [s])^2 (1+2/5)$$ $$=2.84 [mJ]$$

As you can see, about 33% of the total energy was lost, and I don't know where it went. I assume some of it was friction, air resistance, etc., but 33% seems very high for just friction.

This picture is the setup I had for the experiment. $h=8.9 [cm] = 0.089 [m]$, $H = 86.5 [cm] = .865 [m]$, and $x = 0.38 [m]$ (for the marble.) ball path

$\endgroup$
  • $\begingroup$ What is f in your velocity formula? $\endgroup$ – Kalpak Gupta Dec 19 '16 at 4:53
  • 1
    $\begingroup$ What are your measurement errors? 30% error in a variable which depends on the square of a measured value seems entirely plausible for a lab exercise. In my first lab exercise the result I got was g=(5+-15)m/s $\endgroup$ – Christian Vögl Dec 19 '16 at 5:06
  • 1
    $\begingroup$ Your question has too many missing details and is not completely clear. When you say the kinetic energy "just before it hits the ground" you mean at the instant just before the ball reaches its position shown by the 2nd picture of the ball in your diagram, right? If so, then you've neglected the potential energy associated with the distance H. Also, it's not clear how you determined the velocity just before impact. You write that it is 0.38m/0.419sec, but where did these numbers come from and how did you measure them and why is their ratio equal to the instantaneous speed before impact? $\endgroup$ – user93237 Dec 19 '16 at 6:51
0
$\begingroup$

The f stands for the rotational component of kinetic energy, which is 1/2I(w^2), where I is the moment of inertia for a solid sphere, which is 2/5m(r^2). Since there is a '1/2' at the beginning of the line, the 'f' in the formula is '2/5'.

You have not taken into account the vertical translational energy in the parabolic trajectory after it leaves the table.

When you calculate the horizontal speed as 0.38m divided by the 0.419s (I assume) taken, then this is an average velocity, and in the horizontal direction, to assume that this is the same as the final velocity is probably ok (since air resistance will be low).

However, in the vertical direction the final velocity will be twice the average velocity (again assuming low air resistance). This will be where a lot of your missing energy has gone to.

Finally, since the marble drops a total distance of 'h + H', the potential energy available is m.g.(h+H), which will make the starting potential energy calculation a lot higher, and your missing energy even higher !

OK, for an object falling from zero vertical velocity, then the time taken to fall through 0.865m under gravity will be given by the usual s = ut + 1/2a(t^2) formula, which gives a time of: sqrt(0.865*2/9.81) = 0.420s. This tallies with your 0.419s measured.

And so ... the energy released from the fall in the marble's position in the gravity field is given by: m.g.(h+H) = 0.0049*9.81*(0.089+0.865) = 45.86mJ.

... the rotational kinetic energy is given by k.e(rot.) = 1/2.m.(v^2).2/5, and assuming negligible loss in horizontal velocity due to air resistance over the fall to the ground, this is: = 0.5*0.0049*((0.38/0.419)^2)*0.4 = 0.806mJ.

... the translational kinetic energy at the end of the fall is given by using a distance travelled which is the diagonal magnitude of a right-angled triangle with sides 0.38m and 2*0.865m (the 0.865 is doubled because the instantaneous velocity will be twice the average), which is: sqrt((1.73^2)+(0.38^2)) = 1.771m. And so the kinetic energy is the usual 1/2.m.(v^2) = 0.5*.0049*((1.771/0.419)^2) = 43.770mJ.

So, compared to a potential energy release calculation of 45.86mJ, the kinetic energy gain is 43.770 + 0.806 = 44.576mJ.

The difference is 1.284mJ or 2.80% of the potential energy released ... which is close to a more realistic figure for a loss of energy due to friction on the incline and air resistance on the incline and on the downward fall.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ JerryFrog, the 0.419 s is going to be the time that it takes for the marble to leave the table and hit the floor. Note that the OP has several problems with this question: 1) mixed units; 2) very poor layout on a conservation of energy problem; 3) moderate to large uncertainty in obtaining measurements that will give a low % error; 4) conceptual issues whereby a good deal of the gravitational potential energy in the problem was left out of the problem solution. $\endgroup$ – David White Dec 19 '16 at 7:03
  • $\begingroup$ @David White 1) By mixed units, I'm assuming you mean g instead of kg and mJ instead of J... I did this because my teacher told my class to, since it would make it easier to read the results. 2) My bad, I didn't realize I formatted this incorrectly. How would I improve it? 3) The only thing I actually measured was x; the rest was calculated. I can be reasonably sure my x measurement was accurate because it was precise over several trials. 4) Leaving out vertical energy doesn't make sense to me either, but the way I showed is the way my teacher showed my class how to do the lab. $\endgroup$ – Shannon Dec 19 '16 at 11:10
  • $\begingroup$ @Shannon, either the teacher explained it incorrectly, or you misunderstood the directions. Talk to the teacher about this to clear up the confusion. $\endgroup$ – David White Dec 19 '16 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.