
Jared J. answered 01/10/20
Engineering Physics Grad with Prior Math Tutoring Experience
a) This equation here is one of a few kinematic equations (kinematic, meaning related to motion) and it's one of the most used equations in introductory physics courses:
- xf – xi = vi • t + 1/2 a • t 2
- xf is the location of the cylinder at the bottom of the ramp, or its final coordinate
- xi is the starting spot of the cylinder on the ramp, or its initial location
- » Often, the left side of the equation: (xf – xi) is simply called Δx for displacement (the difference, in length, between where the object started and where it ended up).
- vi is the object's initial velocity. We can assume that, in this case, it is 0, meaning that the students just let the cylinder roll down without being pushed.
- t is the time variable. At the moment the cylinder is let go, t = 0
- Finally, a is the acceleration (the value we want to know) that will be calculated from the measured t from the stopwatch and Δx displacement of the cylinder.
In your problem, Δx length difference between the cylinder's start and the bottom of the ramp.
b)
Step 0 is to put the calipers away and to use the clock only to see when your class gets out. The calipers, although more precise, are too short a measuring tool to be useful. The clock is too imprecise.
- Measure the length of the ramp with the meter stick. This length is Δxramp
- Get one student to hold the cylinder at the top of the ramp while a second prepares the stopwatch.
- Start the stopwatch when Student #1 lets go of the cylinder and stop it when the cylinder is observed to hit the bottom of the ramp. Record the time as t
- Repeat steps 2-3 starting the cylinder at 8 different locations up the ramp besides the top, recording Δx and t for each trial (Note: I chose 8 to give us enough data points for part c)
Hopefully this provides enough for the how-to
c)
We're going to need a different equation here because Δx = 1/2 a • t 2 gives us a parabola graph when we put Δx on the y-axis (confusing, I know. But that's actually where it goes): and t on the x-axis SO here a linear kinematic equation:
v = vi + a • t
The v stands for velocity (speed with a direction) at some time t. We know vi = 0 cm/s, we're still looking for a, and we just measured t, but we don't have v yet: only Δx.
Thankfully, we have a relation for v in terms of Δx.
v = Δx / t
For each of the 8 spots you chose on the ramp to roll down the cylinder, plot ( Δx / t ) which equals v on the y-axis and t on the x-axis and you have you straight line. You now know the speed of the cylinder at times t and the slope of that line is a.
d)
Just like v = Δx / t acceleration a = Δv / t. Let's look at the ramp acceleration aramp first: Since the cylinder starts off at rest though we know that Δvramp is just the velocity of the cylinder rolling on the floor vfloor . The time it accelerated is tmeasured , the time you wrote down from your stopwatch.
Now the Δvwall is the velocity difference between the cylinder rolling into the wall, vfloor , and it rolling out, -vfloor . The difference is 2 • vfloor . The time it's in contact with the wall is nonzero, but more importantly it's very small! So short of a time that it's difficult - if not impossible - to measure with a stopwatch. Let's write this out:
Δvramp = vfloor < 2vfloor = Δvwall
tramp = tmeasured > ttoo small to measure = twall
a = Δv / t So awall > aramp
Hope this helps!!
Mbrook E.
Thank you for your outstanding explanations, Jared. I'm helping my son with this same prob in Nov 2020 and came across your work here - which confirms the answers we were coming up with. The time and effort you've put into this are greatly appreciated.11/26/20
Jen H.
Very helpful, thank you!01/10/20