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Problems Slipping and rotation

1 answer  ·  posted 9mo ago by Anonymous‭  ·  edited 1mo ago by Anonymous‭

#6: Post edited by user avatar Anonymous‭ · 2021-12-08T08:54:03Z (about 1 month ago)
  • >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 user avatar Mithrandir24601‭ · 2021-05-05T22:31:08Z (9 months 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 by user avatar Anonymous‭ · 2021-05-01T12:16:52Z (9 months ago)
  • >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 by user avatar Anonymous‭ · 2021-05-01T12:16:16Z (9 months ago)
  • >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 by user avatar Anonymous‭ · 2021-05-01T09:48:43Z (9 months ago)
  • >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 by user avatar Anonymous‭ · 2021-05-01T09:48:17Z (9 months ago)
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