# Rotational motion problem

1. Nov 20, 2008

### bowma166

1. The problem statement, all variables and given/known data
When the wheels of a landing airliner tough the runway, they are not rotating initially. The wheels first slide on the runway (and produc eclouds of smoke and burn marks on the runway, which you may have noticed), until the sliding friction force has accelerated the wheels to the rotational speed required for rolling without slipping. From the following data, calculate how far the wheel of an airliner slips before it begins to roll without slipping: the wheel has a radius of 0.60 m and a mass of 160 kg, the normal force acting on the wheel is 2.0 x 105 N, the speed of the airliner is 200 km/h, and the coefficient of sliding friction for the wheel on the runway is 0.80. Treat the wheel as a uniform disk.

2. Relevant equations
$$\vec{\tau}=\vec{F}\times\vec{R}$$

$$|\tau|=I\alpha$$

$$F_{friction}=\mu_{k}N$$

$$I_{disk}=\frac{1}{2}MR^{2}$$

$$a=\alpha R$$

$$v^{2}=v_{0}^{2}+2ax$$

3. The attempt at a solution
Basically, it's a wheel with a force applied at one point that creates a torque up to a certain acceleration. So... I started with $\tau=RF=I\alpha$, substituted in for I, alpha, and F, and solved for a:
$$R\mu_{k}N=\left(\frac{1}{2}MR^{2}\right)\left(\frac{a}{R}\right)$$

$$\mu_{k}N=\frac{1}{2}Ma$$

$$\frac{2\mu_{k}N}{M}=a$$

I then substituted that value of a into the v2=whatever kinematics equation:

$$v^{2}=v_{0}^{2}+2\left(\frac{2\mu_{k}N}{M}\right)x$$

$$x=\frac{Mv^{2}}{4\mu_{k}N}$$

Plugging in the given values, I get x = .772 m. The book says that the answer is 1.6 m. What am I doing wrong? It is kind of weird that they give you R but I didn't have to use it.

Thanks.

Last edited: Nov 20, 2008
2. Nov 21, 2008

### bowma166

Anyone have any ideas? (Sorry for bumping so early like this but I need to sleep soon and I've got a test tomorrow. Heh.)

3. Nov 21, 2008

### naresh

You're confusing angular displacement of a point on the wheel and the displacement of the airliner. The 'x' you solved for, therefore, is the tangential distance travelled by a point on the wheel, nothing to do with how far the airliner itself moves.

4. Nov 21, 2008

### bowma166

Oh, whoops. Thanks.