Linear velocity problem

Question

The diameter of the back wheel of a tricycle is 1.5 times longer than the front wheel's diameter. When the tricycle is moving, the surface points of the back wheel move with the linear velocity of 4 m/s. Find the linear velocity of the front wheel?

The correct answer is 4 m/s

Answer 1

The key to solving this problem is to figure out the intended meaning of "linear velocity".

I surmise that it is "speed with respect to the center of the wheel", as opposed to with respect to the road.

Different surface points of a wheel move at different speed with respect to the road.

In contrast, they all move with the same speed with respect to center of the wheel.

The statement of the problem uses the plural of "point" in the sentence

"the surface points of the back wheel move with linear velocity of 4 m/s",

and that suggest that they all move with the same velocity.

And that is the property of speed with respect to the center of the wheel.

One argument against my guess is that "velocity" is a vector, which is different

at different points along the perimeter, as the direction of velocity is different.

However, the speeds, i.e. magnitudes of the velocities are all the same.

If you can find out what is meant by "linear velocity" and repost your question, you may get a better answer.

The second question is whether "linear velocity" refers to average velocity of instantaneous velocity.

I do not know, but the solution is the same for both kinds of velocity,

and I will try to explain it independently of which is meant here.

Let me review some general properties of circular motion.

Let

r be the radius of a wheel,

ω be the angular velocity of the wheel,

Δt be a time period.

During that time period Δt the wheel moves by angle ω×Δt.

The arc length of the angle ω×Δt along the perimeter of the wheel is ω×Δt×r.

That is also the distance of the road the wheel coved in time Δt.

The speed of a point on the perimeter (with respect to the center of the wheel) is

ω×Δt×r/Δt = ω×r

Now let's talk about our tricycle whose front wheel has

radius r₁ and rotates with angular velocity ω₁.

In the meantime the back wheel with

radius r₂ rotates with angular velocity ω₂.

In any time period Δt, one wheel covers the distance ω₁×Δt×r₁ and the other covers ω₂×Δt×r₂.

Since the wheels are attached to the same tricycle, those two distances must be the same:

ω₁×Δt×r₁ = ω₂×Δt×r₂

This simplifies into

ω₁×r₁ = ω₂×r₂

And that says that that the two linear velocities are the same.

Answer 2

I understand that very often for motion in one direction along a straight line we interchangeably use the words speed and velocity. But frankly speaking the speed (scalars) is magnitude of velocity (vector).

We will consider here only the case when the wheel of the tricycle rolls smoothly, i.e. the object rolls without slipping. The point of the wheel that touching the road surface has linear velocity of the center of mass of the wheel so that this point all the time is directly above the center of the wheel. Axes of the back wheel that coincides with the center of mass of the back wheel is attached to the tricycle. It means that the center of mass of all wheels (including the front) and the tricycle itself move with the linear velocity whose magnitude is 4 m/s.

The answer to the problem 4 m/s.