# SAVE ME from this weird mechanics problem.

1. Oct 14, 2004

### saksham

Two persons of different mass roll down the same hill from the same starting point on identical bicycles.
Neglect friction and air resistance and assume that the tires roll without slipping. Which person will reach the
bottom first? Prove.

I have mid term tomorrow.

Best,
Saksham

2. Oct 14, 2004

### Gokul43201

Staff Emeritus

Anyway, what ideas do you have ? What equations may be relevant here ?

3. Oct 14, 2004

### saksham

This is my Physics assignment. I am sure that there should be some theory behind it. As you can see that it is either true or false. We should consider rotational and translational kinetic energy.
But who reaches the bottom first?

4. Oct 14, 2004

### Romperstomper

You should start out by drawing the free-body diagrams. The person who gets down the hill the fastest accelerated faster, so you want to find the acceleration of each person.

5. Oct 14, 2004

### Supaiku

It's really the same as dropping two people from the the Empire State Building when one weighs more than the other (of course neglecting air resistance, like your question). Who hits first?
I hope you know, cuz I don't want you to have to do the exeriment
(I would hope you know since this is brought up all the time - I'v probablly heard a version of it fifty times, and 47 of them weren't even in physics class.)

6. Oct 15, 2004

### aekanshchumber

If all the conditions are obeyed then both will reach at the bottom simultaneuosly, distance coverd is same acceleration is same nofriction to oppose the motion therefor neither size nor mass willeffect it. Try equations yourself.

7. Oct 15, 2004

### shashank010288

Since they r of diff masses , normal F will be diff. But since they r rolling friction is sufficient, ( By neglect friction, they mean ROLLING FR.)
SO, THEY WILL REACH SIMULTANEOUSLY.

8. Oct 15, 2004

### ehild

Neglecting friction means that there is no kinetic friction. No slipping means that the static friction is enough to ensure roll. If a wheel rolls without slipping, the speed of its centre of mass is $$v=R\omega$$. Static friction does not do any work, you can use conservation of energy to solve the problem. The potential energy at the top equals to the kinetic energy at the bottom. The kinetic energy is the sum of the translational and rotational ones. There is rotational kinetic energy because of the wheels.
$$KE_{rot}=\frac{1}{2}I\omega^2$$.

Can you continue from here?

ehild

9. Oct 15, 2004

### saksham

Thanks ehild. That was what I wanted but I think we get the result that they both reach the bottom at the same time.
Thanks to all the people who replied.
And, I did not have that question in the midterm, had I got that then I would have got that correct.
Thanks once again.
Cheers,
Saksham

10. Oct 16, 2004

### ehild

You are right if the moment of inertia of the wheels can be neglected. You are wrong if it counts.

ehild

11. Oct 16, 2004

### Staff: Mentor

If you model the bike as having a wheel of mass m, with all its mass on its rim, then you should be able to show that the speed of the bike is independent of the mass of the rider. (Thus they reach the bottom simultaneously.)

Last edited: Oct 16, 2004
12. Oct 16, 2004

### ehild

I may do something wrong, check my derivation, please.

m is the mass of the two wheels. M is the mass of the rider and the other parts of the bike. R is the radius of the wheels, I is the moment of inertia of the two wheels. If all the mass of the wheels is concentrated on the rims then $$I=mR^2$$. If the wheels do not slip the angular velocity of the wheels multiplied by R is the same "v" as the velocity of the centre of mass of the whole bike and rider together.

The bike starts from rest at the top of the slope. The angle of incline is $$\alpha$$.
According to conservation of mechanical energy, the sum of change of KE after travelling a distance s along the incline and the change of PE equals zero.

$$\frac{1}{2}[(m+M)+I/R^2]v^2 = (m+M)gs\sin(\alpha)$$

$$v=\sqrt{ \frac{2(m+M)gs\sin(\alpha)}{m+M+I/R^2}} = ds/dt$$

We get s(t) by integration:

$$s=\frac{(m+M)g\sin(\alpha)}{2(m+M+I/R^2)}t^2$$

If all the mass m is on the rims then I/R^2=m and

$$s=\frac{(m+M)g\sin(\alpha)}{2(m+M+m)}t^2=\frac{g\sin(\alpha)}{2(1+\frac{m}{m+M})}t^2$$

If m is comparable to M the time needed to rich the bottom of the slope does depend on the mass of the rider.

ehild

13. Oct 16, 2004

### decamij

wouldn't heavy person reach first because since they are rolling down a hill, the "forward" component of the force of gravity on the heavier biker will be greater. Therefore, he will accelerate faster. wouldn't he?

14. Oct 16, 2004

### Staff: Mentor

You are absolutely correct. I made an error in my derivation. (I left out the translational KE of the wheels. Oops!) Nicely done, as usual.

15. Oct 16, 2004

### Staff: Mentor

Since the force of gravity is exactly proportional to the mass, that argument would lead to the same acceleration for each.