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Rolling a ball down an inclined slope. Something's wrong.

  • Thread starter Enlightnd
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  • #1
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Hi guys,

Homework Statement


I'm trying to find the theoretical time it takes for a glue stick to roll down an inclined plane of 10 degrees. We then have to compare it to the practical value, which was 1.09 secs.(Average over 10 repetitions). The glue stick has a mass of 10g. The distance of the inclined plane was 41 cm.


Homework Equations


a = g * sin( theta ) for acceleration where theta is 10 degrees.
t = SQRT (2d/a) derived from d = d0 + v0t + (1/2)at2

The Attempt at a Solution


a = 9.8 * sin(10)
a ≈1.70m/s^2
t = SQRT (2* 41/1.7)
t ≈ 6.95.

6.95 does not equal 1.09 at all, so I'm asking did I use the right formula? If not what are the right formulas. And could someone please provide me with a worked example so I can understand this more clearly.

I'm only a high school student doing this as an independent research project so please don't use advanced language, if you have to can you please explain it as well.

Thank you very much, you help is much appreciated. :D
 

Answers and Replies

  • #2
28
3
Here you have only taken into account the translational motion of glue stick.Glue stick is like a small cylinder.It will rotate as well as slide on the inclined plane(given that the inclined plane has friction).
If you don't want to take into account the rotational motion,then also you are missing one thing i.e.friction force(=umgcos(theta)).It acts opposite to motion Hence
a=g (sin(theta)-ucos(theta))..(u=coefficient of friction).

If you take into account rotational motion of object,the equation of accn will be a bit more complex.

The answer which you found will be true only in case of frictionless incline which is not possible in real life.
If any more problems,Hit me up again
 
  • #3
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Thank you very much for your reply. I see that I'm missing friction. So how would i find out the friction between the gluestick and inclined plane? I didn't really understand the last few sentences when you mentioned the formula's relating to friction and acceleration.

*edit. OK I understand some bits now. I've got theta and cos and sin. So the only thing i need now is (u) providing that I don't take into account slipping and sliding. How would i find out (u) then?
 
  • #4
28
3
No problem Buddy.Every two pair of objects has a particular value of coefficient of fricition(u) which is determined by experiments.Frictional force =uN (u is the coefficient of friction and N is the normal acting on the object).Just asking,Have you read the chapter Laws of motion?
In case of inclined plane (N=mgcos(theta)).so friction=umgcos(theta).
Two forces act on the body mgsin(theta) and umgcos(theta) which are opposte in direction..
F(net)=mgsin(theta)-umgcos(theta)
m.a(net)=mgsin(theta)-umgcos(theta)
a(net)=gsin(theta)-ugcos(theta)
 
  • #5
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Thank you very much for your reply. I see that I'm missing friction. So how would i find out the friction between the gluestick and inclined plane? I didn't really understand the last few sentences when you mentioned the formula's relating to friction and acceleration.

*edit. OK I understand some bits now. I've got theta and cos and sin. So the only thing i need now is (u) providing that I don't take into account slipping and sliding. How would i find out (u) then?
You can only find it out by doing experiments if you beforehand don't know it..Make you clear that the cylinder does slide here .We have only assumed that the cylinder is not rotating..

Have you read Laws of Motion?IF not,then leave it.
 
  • #6
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Thank you again for your reply. After searching the internet, it said that the coefficient of Friction (u) regarding plastic on wood, was 0.5 and 0.4. Do i just substitute these values in the formula that you provided me with in your first response? And no I haven't read the chapter Laws of motion, although I am looking forward to doing physics in senior high school. Also, regarding 0.5 and 0.4, which value do i substitute? Thank you very much.
 
  • #7
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Oh. But the cylinder here is rotating. Does that mean I have to use another formula regarding this project? :((
 
  • #8
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I really appreciate your experimental approach but you need to study a bit more to do it properly..When the cylinder rotate,the equation of the acceleration are quite complex.So don't bother about it..My suggestion would be to take a small point object and experimentally determine the time taken by it to reach the bottom of inclined plane.
Then calculate the theorotical accn using the equation I provided above and then the theorotical time.There are two values of coefficent of friction for a particular pair.Always take the one with the smaller value.And no need to thank me every time.Have fun and do take Physics.It's aweosme
Are you in Class 10?..
 
  • #9
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One thing Princu failed to point out to you is that you made a mistake with units in your problem. 41 cm is 0.41 meters, so your predicted time to go down the plane should have been 0.695 sec rather than 6.95 seconds.

Chet
 
  • #10
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One thing Princu failed to point out to you is that you made a mistake with units in your problem. 41 cm is 0.41 meters, so your predicted time to go down the plane should have been 0.695 sec rather than 6.95 seconds.

Chet
So sorry, didn't see that.
 
  • #11
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I really appreciate your experimental approach but you need to study a bit more to do it properly..When the cylinder rotate,the equation of the acceleration are quite complex.So don't bother about it..My suggestion would be to take a small point object and experimentally determine the time taken by it to reach the bottom of inclined plane.
Then calculate the theorotical accn using the equation I provided above and then the theorotical time.There are two values of coefficent of friction for a particular pair.Always take the one with the smaller value.And no need to thank me every time.Have fun and do take Physics.It's aweosme
Are you in Class 10?..
Uhmmm, no I'm in Year 9 in a selective high school in Sydney Australia, so maybe class 9? I don't know.
 
  • #12
28
3
One thing Princu failed to point out to you is that you made a mistake with units in your problem. 41 cm is 0.41 meters, so your predicted time to go down the plane should have been 0.695 sec rather than 6.95 seconds.

Chet
Thanks for correcting. By the way I did succeeded I pointing out the rest.Ha Ha:wink:
 
  • #13
28
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Uhmmm, no I'm in Year 9 in a selective high school in Sydney Australia, so maybe class 9? I don't know.
Very well.Go ahead and God bless.I have completed my school education and currently preparing for engineering entrance exams due next year.
 
  • #14
jambaugh
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The problem here is that the potential energy due to gravity is being converted both into kinetic energy of the linear motion, and of the rotational motion since the object is rolling.

There's a way to reduce this down to an "effective mass" for the inertial mass of the system. A solid cylinder which has a moment of inertia of [itex]I=1/2 m r^2[/itex]. The kinetic energy of the cylinder as it rolls is the rotational plus linear K.E.

[tex]T = \frac{1}{2} m V^2 + \frac{1}{2} I \omega ^2 [/tex]

Since as it rolls it rotates so that [itex]V=r \omega[/itex], we get...

[tex]T = 1/2 m V^2 + 1/4 m r^2 * V^2 / r^2 = ... = \frac{1}{2}( 1.5m) V^2=\frac{1}{2} M_{eff}V^2[/tex]

OK so the effective mass [itex]M_{eff}[/itex] is 1.5 times the physical mass... this is because there's more kinetic energy tied up in the rolling cylinder than would be a sliding one. Now this extra effective mass doesn't contribute to the weight, i.e. the force due to gravity so you'll see the acceleration have a factor of mass/effective mass or 2/3 the value you used.

Carrying this factor through to the time you'll get a [itex] \sqrt{3/2} \approx 1.225[/itex] correction factor to the time calculates assuming sliding with no friction. In other words for a given problem a solid cylinder will take about 1.225 times longer to roll down than a frictionless object would to slide.

This assumes the chapstick is a solid cylinder of uniform density.

If we go to the extreme and consider a hollow cylinder rolling down a slope you get an effective mass of 2m and so a time correction of [itex]\sqrt{2}\approx 1.414[/itex]. In the case where all the mass is concentrated at the axis of the cylinder the effective mass is 1 m and the cylinder behaves as a point mass sliding down without friction.
 
  • #15
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Because your cylinder is rotating as it descends, some of the energy goes into spinning it up. So its overall motion is slower. Have you studied rotational motion? Moment of inertia ring a bell?
 
  • #16
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Uhh, to be completely honest, I have not studied anything you guys have mentioned. So yeh I'm a complete noob in physics. Its only an optional extension in our independent research project.
 
  • #17
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Then I am afraid you can only do the lower bound estimate that you did (but mind the units as Chet said) and then do the hand-waving argument that "rotation slows it down".
 
  • #18
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Ahaha. Ok then. Thanks anyways.
 
  • #19
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Jambaugh really gave you the result you need. According to what he presented, the frictional force is equal to ma/2 for a cylinder of uniform density. So, as he said, the time should be about 1.225 times as long. 1.225 x 0.695 = 0.85 sec. This is close to your measured value of 1.09 seconds, and is about the best your theoretical estimate is going to do.
 

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