palaphys
- 269
- 19
- Homework Statement
- A ring of mass m and radius r is rotating on a horizontal rough table with coefficient of friction. how long does the ring take to stop rotating? What is the frictional force from the ground?
- Relevant Equations
- Torque=r×F
So to begin with, I did this:
$$
\begin{align*}
dm &= \frac{M}{2\pi} d\theta \\
dN &= dm \cdot g = \frac{M}{2\pi} g \, d\theta \\
df &= \mu_k \, dN = \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \int df = \int_0^{2\pi} \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \mu_k \frac{M}{2\pi} g \int_0^{2\pi} d\theta \\
f &= \mu_k \frac{M}{2\pi} g \cdot 2\pi \\
f &= \mu_k M g
\end{align*}
$$
However, I had a thought- what if I could simply super impose all the frictional force vectors on every small element? Wouldnt that add up to 0?As there are vectors of equal magnitude in every direction.. Then why do I end up with some other answer?
$$
\begin{align*}
dm &= \frac{M}{2\pi} d\theta \\
dN &= dm \cdot g = \frac{M}{2\pi} g \, d\theta \\
df &= \mu_k \, dN = \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \int df = \int_0^{2\pi} \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \mu_k \frac{M}{2\pi} g \int_0^{2\pi} d\theta \\
f &= \mu_k \frac{M}{2\pi} g \cdot 2\pi \\
f &= \mu_k M g
\end{align*}
$$
However, I had a thought- what if I could simply super impose all the frictional force vectors on every small element? Wouldnt that add up to 0?As there are vectors of equal magnitude in every direction.. Then why do I end up with some other answer?
