Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs

Dashboard
Notifications
Mark all as read
Problems

Slipping and rotation

+1
−0

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?

$$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

1 comment thread

General comments (1 comment)

1 answer

+1
−0

I know this is a bit late, but if you still haven't figured it out, here goes:

Firstly, let the mass per unit area of the disk be $\sigma = \frac{m}{\pi R^2}$.

Consider a small element of area $dA$ at a distance $x$ from the center of the disk. The mass of this element is $dm = \sigma dA$. The frictional force on this element is $dF = \mu (dm) g = \mu \sigma g \times dA$

The torque on this small element is $d\tau = x \times dF = \mu \sigma g x \times dA$. (note that by $\times$ I just mean normal multiplication, not the cross product or anything).

Now let us consider a thin circular ring with thickness $dx$ at a distance $x$ from the center (I mean to say a ring of radius $x$ whose center is the center of the disk). The net torque $\tau$ on this ring would be the integral of all the small torques on the small elements of area $dA$ lying on this thin ring.

$$\tau = \int d\tau = \mu \sigma g x \int dA = \mu \sigma g \times (2 \pi x^2 dx)$$

Note that in this integral $x$ is a constant, since $x$ is the distance from the center for all the small elements on the circular ring we have taken under consideration.

As $dx \to 0$, we have infinite such rings on the disk. So now we need to integrate over all these rings. The equation above matches the second equation of your post.

I think you should be able to take it from here. The important thing is to keep in mind what we are doing: first we find the net torque on a thin ring situated at a distance $x$ from the center which I've done above, and then we integrate over all such rings.

Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us! Donate