Speed Down An Incline

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

MedPR

Membership Revoked
Removed
10+ Year Member
Joined
Dec 1, 2011
Messages
18,577
Reaction score
57
Which graph best represents the speed of a rolling mass as a function of the distance it travels down the inclined plane?

RrhmJ.jpg


I say graph 1, TBR says graph 2. Ignore all the crooked lines and stuff. Just consider the overall shape 🙂

As a mass rolls down an incline, it picks up speed according to a=gsintheta. So, like in free fall, every time it moves a unit of distance, it will be going faster than the previous...

Nevermind, I just figured out what I was confused about....

But now I'm confused about something else. If you roll a ball down an incline, why would the speed every stop increasing linearly?

Members don't see this ad.
 
I would probably use v^2 = 2ax for this problem, as we start from an initial speed of 0m/s. a is some constant less than g, depending upon the angle of the ramp. As the graph is displaying v(d), the original equation can be modified to v = (2ax)^(1/2), so v is proportional to the square root of x or d, so that would correspond to graph 2. Graph 1 shows the inverse relationship, or in other words, if only the axes labels were switched, graph 1 would be correct.
 
But now I'm confused about something else. If you roll a ball down an incline, why would the speed every stop increasing linearly?

Just like something falling through the air, it's reached terminal velocity.
 
Members don't see this ad :)
Just like something falling through the air, it's reached terminal velocity.

--wrong--
That. You need some friction between the rolling object and the ramp for it to rotate. And if you have friction, you will have terminal velocity.
--wrong--

No terminal velocity in this case, see later post.
 
Last edited:
I immediately jumped to terminal velocity, but I thought that typically, as another source says it "terminal Velocity is only a possibility when you are dealing with fluid friction (air is a fluid) as opposed to contact friction like static or kinetic friction."

Is the term really "terminal velocity"
 
Disregard what I said, it's incorrect. SaintJude is right, there is no terminal velocity here. The key is that the graph is velocity vs distance, not velocity vs time.

The easiest way to see how they depend on each other is from the formula that v = (2ax)^(1/2) posted.

A qualitative way to interpret that is that as the speed increases, the distance per interval of time increases while the change in speed does not. In other words ΔV/t stays the same but Δd/t increases.

A major difference between this graph and the terminal velocity graph is that the terminal velocity graph has an asymptote for the velocity which limits at a certain number, in this problem the velocity increases to infinity.
 
Top