Rolling Wheel on Inclined

Rolling wheel

A full wheel rolls on an inclined plane without slipping. In an experiment, you measure the following experimental data for this rolling wheel:

Wheel Radius: $R = (3.25 \pm 0.05) cm$

Wheel mass $m = (0.43 \pm 0.01) kg$

Incline angle $\theta = 28 deg\pm 2 deg$

Incline length: $L = (1.48 \pm 0.01) m$

Time to roll down  $t = (1.04 \pm 0.02) s$

a. Usually the wheel inertia moment is $I = k m R^2$. Please find $k$ value and its uncertainty.

b. Please find $I$ and its uncertainty.

You can find an entire lesson about the moment of inertia here.

rolling wheel inertia

Two rotating spinners. Their moment of inertia keep them rotating.

Inertia moment and uncertainty

The acceleration on the incline is $a=2L/t^2$ with an uncertainty of

$d a=2(d L/t^2)+2(L/t^3)d t$ so that

$a=2*(1.48/1.04^2)=2.7367 m/s^2$ and $d a=0.071 m/s^2$

(the uncertainty of a value is found by differentiating it and changing all “minus” signs to “plus” because all the partial uncertainties add together).

For example for the equation $F=m*a$ its differential is $dF=a*dm+m*da$. Thus the uncertainty in force depends on the uncertainties in mass and acceleration (each multiplied with the other value in the equation) added together.

Final speed is (from the uniform motion equation with initial speed zero): $v=at=2.7367*1.04=2.8462 m/s$

and thus the speed uncertainty is

$d v=t*d a+a*d t=0.1286 m/s$

One can find the acceleration of the wheel on the incline from energy considerations.

Initial potential energy=final rotation energy (kinetic rotation) +final translation energy (kinetic translation)

$\frac{m g L}{\sin\theta}=\frac{I\omega^2}{2}+\frac{m v^2}{2}$    or  $\frac{m g L}{\sin\theta}=\frac{v^2}{2}(\frac{I}{R^2}+m)$  or  $\frac{m g L}{\sin\theta}=\frac{m v^2}{2}(k+1)$  or $k=\frac{2gL}{v^2\sin\theta}-1=60.85$

Therefore the uncertainty in k is

$d k=\frac{2g*d L}{v^2\sin\theta}+\frac{4gL*d v}{v^3\sin\theta}+\frac{2gL*d\theta}{v^2\cos\theta}=8.88$

So that the moment of inertia of this rolling wheel is $I=k m R^2=60.85*0.43*0.0325^2=0.0276 kg*m^2$

with an uncertainty having of $d I=m R^2*d k+k R^2*d m+2kmR*d R=0.00633 kg*m^2$