- #1

seraphimhouse

- 28

- 0

## Homework Statement

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

## Homework Equations

tnet = I(alpha)

at [tangential acceleration] = (alpha)r

## 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**

[alpha] = 4.18 rad/s^2

[alpha] = 4.18 rad/s^2

Yet, I got the answer wrong. Please help! I've been staring at this problem for too long.