# Speed of point on a wheel

• Aug 22nd 2008, 12:38 PM
Aslak
Speed of point on a wheel
A bicycle moves at 10 km/h on a horisontal road. How fast does the foremost point on the wheel move?

The answer is supposed to be 14 km/h, but I haven't got a clue where to look for it...A little help would be appreciated. (Happy)
• Aug 22nd 2008, 01:15 PM
arbolis
I'm not sure, but I think we are missing some data. What about the radius of the wheel?
• Aug 22nd 2008, 01:51 PM
Aslak
That's what I thought too...No that's all the information we get; guess it's unsolvable then.^^

Anyhoo; if we set radius=1. How do you calculate the speed of the point? I've never done something like that either. =)
• Aug 22nd 2008, 01:55 PM
arbolis
Quote:

Anyhoo; if we set radius=1. How do you calculate the speed of the point? I've never done something like that either. =)
Me neither. I guess you are asking the speed of a point on a wheel with respect to an observator situated on the road and not moving at all compared to a point on the road. If so, I'll have to think a bit. I hope someone will come up with something we are probably missing. Also, you said we put the radius=1, but 1 what? Meter? That make a very big bike!
EDIT : I also believe that the speed will vary in function of the position (which vary along the time) of the point we are considering. So I guess I take the maximum...
EDIT2: Nevermind. My intuition will lead me nowhere, I should start to write down something.
EDIT 3: Well, I got that the velocity of the point with respect of the man on the bicycle is about 2.78 m/s. But this velocity changes in direction. It appears that sometimes it has the same direction than a parallel line to the floor. Maybe the velocity of the point would be 12.78 m/s as a maximum for anyone sitting on the road watching the point on the bicycle, but I'm not 100% sure. If someone confirm this, then I'll post how I found that.
• Aug 22nd 2008, 02:14 PM
Aslak
I see your points. Maybe we should change the conditions a bit then. What about:

A big bicycle is moving at 5 m/s. The radius of the wheel is 1 metre. How fast is a point on the bike moving relative to the road?

Maybe periodic functions come into play here?

Edit: Forget the new question if you'd like. Seems like you've almost solved my first. =)
• Aug 22nd 2008, 02:17 PM
arbolis
Quote:

Originally Posted by Aslak
I see your points. Maybe we should change the conditions a bit then. What about:

A big bicycle is moving at 5 m/s. The radius of the wheel is 1 metre. How fast is a point on the bike moving relative to the road?

You're probably right; functions come into play here.

That's ok, but I just did some calculus. If someone confirm them, I'll tell you how I found my answer, and you'll do the same for your new problem. (Wink)
• Aug 22nd 2008, 02:25 PM
nikhil
Quote:

Originally Posted by Aslak
A bicycle moves at 10 km/h on a horisontal road. How fast does the foremost point on the wheel move?

The answer is supposed to be 14 km/h, but I haven't got a clue where to look for it...A little help would be appreciated. (Happy)

Thats a high school question??? wow!!
Ok the answer is beyond my reach this question purely belongs to physics but I can put some light on the concept.
When a bicycle moves its wheel has two types of velocities
1)linear(that is the velocity of center of mass of the bicycle)
2) rotational
instantly the part that touches the ground has velocity=0.
The foremost part which is at the horizontal level of axle has a velocity =the linear velocity of bicycle(that is the velocity of its center of mass) and the uppermost part of wheel has a velocity that is twice of the linear velocity of the bicycle(that is the velocity of its center of mass)
well are you supposed to take physics under consideration?
May be thats beyond high school
• Aug 22nd 2008, 02:34 PM
arbolis
Quote:

Ok the answer is beyond my reach also there is no need of radius as this question purely belongs to physics but I can put some light on the concept.
When a bicycle moves its wheel has two types of velocities
1)linear(that is the velocity of center of mass of the bicycle)
2) rotational
instantly the part that touches the ground has velocity=0.
The foremost part which is at the horizontal level of axle has a velocity =the linear velocity of bicycle(that is the velocity of its center of mass) and the uppermost part of wheel has a velocity that is twice of the linear velocity of the bicycle(that is the velocity of its center of mass)
If your answer is 14km/h then locating that point is actually not so easy.
No need of the radius? If you look at the formula of the linear speed with respect to the rotational speed, then you'll see the radius is very important. But I might have misunderstood the problem. What mean exactly the "foremost point on the wheel"? If it is any point that touch the ground at every turn of the wheel, then its velocity is not 0 because when the rotational acceleration is constant (like in this case), then the modulus of the velocity doesn't change. So it cannot be 0 m/s when touching the ground and then suddenly growing up...
• Aug 22nd 2008, 02:46 PM
nikhil
Quote:

Originally Posted by arbolis
No need of the radius? If you look at the formula of the linear speed with respect to the rotational speed, then you'll see the radius is very important. But I might have misunderstood the problem. What mean exactly the "foremost point on the wheel"? If it is any point that touch the ground at every turn of the wheel, then its velocity is not 0 because when the rotational acceleration is constant (like in this case), then the modulus of the velocity doesn't change. So it cannot be 0 m/s when touching the ground and then suddenly growing up...

Except for the 3 points that I have mensioned radius might be required (thats why I already edited the post for clarification)
2) I said INSTANTLY the part that touces the ground.if its not 0m/s then it will become a case of rolling.
• Aug 22nd 2008, 02:56 PM
arbolis
Quote:

2) I said INSTANTLY the part that touces the ground.if its not 0m/s then it will become a case of rolling.
You're right! I never think about that. (Notice also that it requires a frictional force between the wheel and the ground, otherwise it wouldn't be possible to make the velocity of the touching point equal to 0 m/s.)
• Aug 22nd 2008, 03:03 PM
nikhil
Quote:

Originally Posted by arbolis
You're right! I never think about that. (Notice also that it requires a frictional force between the wheel and the ground, otherwise it wouldn't be possible to make the velocity of the touching point equal to 0 m/s.)

Yupp friction is important for sure otherwise it will become a case of PURE ROLLING.
Well me also counting on topsquark for complete clarification.
• Aug 22nd 2008, 03:19 PM
skeeter
the question asks "how fast", which one can interpret as speed, and adds to the confusion. the question should ask what is the velocity of the foremost point on the wheel.

every point on the wheel has a translational component and a rotational component of velocity. the overall velocity at any point on the wheel is the vector sum of these two components.

the translational component is the same for every point on the wheel ...
10 km/hr in the direction that the bike travels.

the rotational component is tangent to the direction of rotation, and on the outer edge of the wheel, has a magnitude of 10 km/hr, but continuously changes direction.

since the foremost edge of the tire has a rotational component of 10 km/hr tangent to the wheel, the rotational component points straight down, perpendicular to the translational direction.

so ... the two components have identical magnitudes of 10 km/hr and are perpendicular to each other at the point in question. the resultant vector has a magnitude of $10\sqrt{2}$ km/hr, or about 14 km/hr, 45 degrees below the horizontal.
• Aug 23rd 2008, 01:03 AM
Aslak
Sorry for the confusion guys. I don't really know the difference between speed and velocity...
So the radius doesn't matter then? Interesting...

So for my new question with the bicycle with velocity=5 m/s, the magnitude of the resultant vector of the foremost point on the wheel equals:

$\sqrt{5^2+5^2}=5 \cdot \sqrt{2}$ m/s with direction 45 degrees below the horisontal plane.

Some new things really clicked into place here. (Rofl)