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A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long ...
#6: Post edited
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- \begin{align*}
- \mathrm{df} &=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx} \\\
- \implies \mathrm{d} \tau &=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}} \\\\
- \implies\int \mathrm{d} \tau &=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x}
- \end{align*}
- Again, \begin{align*}
- \tau &=\frac{2}{3} \mu \mathrm{mgR} \\\
- \tau &=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha \\\
- \implies \alpha &=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}} \\\\
- \alpha \mathrm{t}=\omega \implies \mathrm{t}&=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- \end{align*}
In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- \begin{align*}
- \mathrm{df} &=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx} \\\
- \implies \mathrm{d} \tau &=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}} \\\\
- \implies\int \mathrm{d} \tau &=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x}
- \end{align*}
- Again, \begin{align*}
- \tau &=\frac{2}{3} \mu \mathrm{mgR} \\\
- \tau &=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha \\\
- \implies \alpha &=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}} \\\\
- \alpha \mathrm{t}=\omega \implies \mathrm{t}&=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- \end{align*}
- Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other way to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
#5: Post edited
by
Mithrandir24601
·
2021-05-05T22:31:08Z (over 3 years ago)
fixed Latex formatting
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
$$\mathrm{df}=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx}\\\Rightarrow \mathrm{d} \tau=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}}\\\Rightarrow \int \mathrm{d} \tau=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \\\text{Again}, \quad \tau=\frac{2}{3} \mu \mathrm{mgR}\\\tau=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha\\\Rightarrow \alpha=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}}\\\alpha \mathrm{t}=\omega \Rightarrow \mathrm{t}=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}$$- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
$$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- \begin{align*}
- \mathrm{df} &=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx} \\\
- \implies \mathrm{d} \tau &=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}} \\\\
- \implies\int \mathrm{d} \tau &=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x}
- \end{align*}
- Again, \begin{align*}
- \tau &=\frac{2}{3} \mu \mathrm{mgR} \\\
- \tau &=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha \\\
- \implies \alpha &=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}} \\\\
- \alpha \mathrm{t}=\omega \implies \mathrm{t}&=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- \end{align*}
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
#4: Post edited
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
>$$- \mathrm{df}=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx}\\
- \Rightarrow \mathrm{d} \tau=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}}\\
- \Rightarrow \int \mathrm{d} \tau=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \\\text{Again}, \quad \tau=\frac{2}{3} \mu \mathrm{mgR}\\
- \tau=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha\\
- \Rightarrow \alpha=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}}\\
- \alpha \mathrm{t}=\omega \Rightarrow \mathrm{t}=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- $$
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- $$
- \mathrm{df}=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx}\\
- \Rightarrow \mathrm{d} \tau=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}}\\
- \Rightarrow \int \mathrm{d} \tau=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \\\text{Again}, \quad \tau=\frac{2}{3} \mu \mathrm{mgR}\\
- \tau=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha\\
- \Rightarrow \alpha=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}}\\
- \alpha \mathrm{t}=\omega \Rightarrow \mathrm{t}=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- $$
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
#3: Post edited
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
[![enter image description here][1]][1]- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
[1]: https://i.stack.imgur.com/Ho1TS.png
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- >$$
- \mathrm{df}=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx}\\
- \Rightarrow \mathrm{d} \tau=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}}\\
- \Rightarrow \int \mathrm{d} \tau=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \\\text{Again}, \quad \tau=\frac{2}{3} \mu \mathrm{mgR}\\
- \tau=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha\\
- \Rightarrow \alpha=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}}\\
- \alpha \mathrm{t}=\omega \Rightarrow \mathrm{t}=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
- $$
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
#2: Post edited
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
Some had solved the process following way- [![enter image description here][1]][1]
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
- [1]: https://i.stack.imgur.com/Ho1TS.png
- >A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
- Someone had solved the problem following way
- [![enter image description here][1]][1]
- In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
- $$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
- [1]: https://i.stack.imgur.com/Ho1TS.png
#1: Initial revision
Slipping and rotation
>A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?
Some had solved the process following way
[![enter image description here][1]][1]
In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?
$$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$
[1]: https://i.stack.imgur.com/Ho1TS.png