0
$\begingroup$

If I have a wheel, I know it's mass and diameter and the slope of a hill. Can I calculate the time it will take to get to the bottom of the hill? I am doing a project for my science fair and I sent 5 wheels down a hill. I thought the biggest wheel would go fastest but it didn't. I don't know why. The middle wheel wheel went the fastest. It was right in the middle of the weight and diameter. The smallest and lightest wheel went the same speed and the heaviest and biggest wheel.

I know the diameter and masses of the wheels. The length of the hill and the degrees of the slope. Is there a formula that can tell me how fast it should go? I did not push it, I let all the wheels roll just by letting go of them so there wasn't any force.

If anyone could help me understand this my Mom will try to explain what you tell me.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

You can calculate it, but some assumptions would have to be made. Namely the wheel does not slip on the surface, air friction can be neglected and the wheel/surface does not deform big inelastic deformations (which would also dissipate energy).

The result depends on de (mass) moment of inertia:

enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ Hi, no matter how many times I roll the wheels down the hill, I get different results each time. It's an 8ft slide and they are skateboard wheels with a hole in the middle. they are the same material. I thought the bigger wheel would win more often but it doesn't. $\endgroup$
    – user44128
    Apr 13, 2014 at 19:35
2
$\begingroup$

You'll need to find the individual moment of inertias of the wheels. If they are hollow cylindrical shells with negligible thickness then you can take their moment of inertia about the rotational axis as $mR^2$ where $m$ is the mass of the wheel and $R$ is the radius.

enter image description here

Now, as fibonatic mentioned, you will have to assume that the wheel does not slip and there is no air drag. Let $\omega$ be the angular velocity at an instant and $v$ be the linear velocity. Hence, $v=\omega R$. Consider the diagram below,

enter image description here

Linear acceleration, $a=\alpha R$

Torque on the wheel will be $$\tau =I\alpha =f_sR$$

So, $$f_s=\frac{I\alpha}{R}=\frac{Ia}{R^2}$$

Also, we can see that $$mgSin\theta - f_s=ma$$

Solve the above equations to get $a$ for each of the wheels.

Improve this answer
$\endgroup$
11
  • $\begingroup$ $a!=g\sin\theta$ $\endgroup$
    – evil999man
    Apr 8, 2014 at 8:23
  • $\begingroup$ use \sin­­­­­­­ $\endgroup$
    – evil999man
    Apr 8, 2014 at 8:23
  • $\begingroup$ @Awesome what do you mean? $\endgroup$
    – user42733
    Apr 8, 2014 at 14:48
  • $\begingroup$ You wrote a= g sin theta which contradicts your last equation $\endgroup$
    – evil999man
    Apr 8, 2014 at 14:57
  • $\begingroup$ @Awesome yeah you are right, thanks for pointing it out, I'll edit it right away. $\endgroup$
    – user42733
    Apr 8, 2014 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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