# Homework Help: Pulled box

1. Nov 29, 2009

### seraphimhouse

1. The problem statement, all variables and given/known data
A wheel of radius 0.358 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.0421 kg·m2. A massless cord wrapped around the wheel is attached to a 3.21 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 4.31 N is applied to the block as shown in Fig. 10-54, what is the angular acceleration of the wheel? Take the clockwise direction to be the negative direction and assume the string does not slip on the wheel.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_54.gif

2. Relevant equations

tnet = I(alpha)

at [tangential acceleration] = (alpha)r

3. The attempt at a solution

I did the forces of the box and got that Ft = p [the horizontal force] - ma where m is the mass of the box.
So, Ft = p - ma

I found the net torque of the disk rotating at the center of its axis to be -Ftr and the rotational inertia of the disk to be I = 1/2Mr^2 where M is the mass of the disk.
So, Tnet = -Ftr & I = 1/2Mr^2

We know that Tnet = I[alpha] so therefore,
-Ftr = 1/2Mr^2[alpha]

We know that the string does not slip on the wheel, therefore the tangential acceleration is equal to the angular acceleration times it's radius.
at = [alpha]r
[alpha] = at/r

So by substituting it to the previous equation, I get
-Ftr = 1/2Mr^2(at/r)
Cleaning up this equation I get
Ft = -1/2Ma

Before I go any further, I found the mass of the disk by using I = 1/2Mr^2. We know that I=0.0421 kg m^2.
0.0421 = 1/2Mr^2 we know what r is so therefore,
M = .657 kg

So I then set the tension found using the disk and the free body diagram of the box equal to each other to get
P - ma = -1/2Ma
cleaning this up and having acceleration set on one side of the equation I get,
a = 2p/[2m-M]
a = 1.496 m/s^2

And finally I substituted acceleration back into the tangential acceleration equation which was
[alpha] = a/r

2. Nov 29, 2009

### Delphi51

Probably a mistake to assume I = .5mr².
You can figure out the rotational part from
torque = Iα
Tr = I*a/r (where T = tension)
without the assumption.
If you sub this expression for T into the T = p - ma you can solve for acceleration and you'll get a smaller value.

3. Nov 29, 2009

### seraphimhouse

I substituted it for T = I$$\alpha$$
I used -Ftr = I [a/r]
Them subbed in Ft = P - ma
rma - rP = I a/r

I cleaned it up and still got 4 m/s^2 which still gave me 20 rad/s^2

4. Nov 29, 2009

### Delphi51

Ah, I didn't have that minus sign. Must rethink the whole thing!
The question specifies counterclockwise as positive, so the first equation should be Ft - p = ma
And the circular one Ft*r = I*a/r
So I*a/r² - p = ma
(I/r² - m) a = p
a = p/(I/r² - m)
a = 4.31/(.0421/.358² - 3.21) = -1.496
so we agree on the number except for the sign. Could the computer be upset over the missing minus sign? It clearly is correct - the p force makes it accelerate clockwise which is defined as negative.