# Homework Help: Determining Velocity Down a Frictionless Incline Plane, Starting at Rest. (BASIC)

1. May 15, 2012

### XADS230

Here's my scenario...I ran these calculations and I feel they are off due to my velocity values.

I have a washer rolling down an incline plane set at 10°.
The Length of the Plane is still unknown.
Starting at rest, determine the velocity at 2". At this distance a camera with take a snapshot and determine whether it's a good part or bad part. The camera takes 0.6 seconds to process the image.
If it's a Bad Part, it will be rejected from the slide(0.6sec later). Determine how far down the incline the washer will be, to know where to reject the washer.
If it's a Good Part, it will continue down the slide where it will checked by another camera, a distance of 2.25" from the first reject spot determined above.
Determine how far down the second reject spot would be on the incline.
Determine the overall length the incline would need to be to complete the mentioned process.

What I know...
Vi = 0.0
Angle = 10°
Frictionless plane.
Constant Acceleration
d1 = 2"
t(camera processing) = 0.6 seconds

Vf^2 = Vi^2 + 2(g*sin(θ)) d1

Plugging in the known values I came up with...

Vf^2 = 0 + 2(385.8in/s^2 * sin(10°))(2")

Yielding me with a result of Vf1 = (16.369in/s).

Over a distance of 2 inches, this velocity seems way too fast. Should d1 actually be the height of the triangle created by using 2" as the hypotenuse and Sin(10°)? Resulting in a new value of Vf1 = (6.819 in/s).

For the Second part of my calculations, I have to determine it's distance based off the velocity and time known. To determine position with no final velocity I used the equation

X = Vi(t)+(0.5)a*t^2

With the Velocity of 16.369in/s and the time value, I calculated a reject distance at 21.88" down the ramp. And repeated the same method of calculations to get the second reject distance and overall length. Can someone confirm my suspicions?

2. May 15, 2012

### nasu

Do you really mean "rolling down an incline" or is just sliding?
If it's just sliding on a frictionless plane, the speed after 2" seems about right. (I get about 11.5 in/s).

3. May 15, 2012

### XADS230

Yes, I mean rolling.

4. May 15, 2012

### XADS230

I just realized that I have to account for the force of gravity center of mass and rotational acceleration of a disk along with friction.

Vf = √(Vi^2+2(g*sin(θ)/1+I/mr^2)d1)

I (in this case: Disk) = 1/2m*r^2

so,

Vf = √(Vi^2+2((2/3)g*sin(θ))L)

5. May 15, 2012

### haruspex

If it's rolling you don't mean frictionless.
Since it's a washer, there'll be a hole in the disc, no? If the internal and external radii are A, B then the M.I. is m(B^2+A^2)/2.
Not sure why you concern yourself with the velocities. Don't you just want to know the times to various distances?
Force down plane = mg.sin(θ) = m.a + [m(B^2+A^2)/2].a/B^2
a = 2g.sin(θ) /[3 + A^2/B^2]
where a is the linear accn.
Then solve 2s = at^2 for the various s.

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