# Disk and Ball on a smooth surface [closed]

A uniform disk of mass $\text{2m}$ and radius $\text{R}$ is placed freely on a smooth surface as shown in the figure. A particle of mass $\text{m}$ is connected to the circumference of the disc with a massless string. Now an impulse $\text{J}$ is applied on the particle in the direction shown by the dotted line. What is the acceleration of center of mass of the disk just after application of impulse?

$\text{Details}$

1) Neglect friction everywhere.

2) $\text{J}=10$, $\text{m}=\sqrt{10}$ and $\text{R}=0.25$.

3)Each of the given quantities are in $\text{S.I.}$ units.

At first glance, it seemed to me a question that uses chaos theory. However, since this is in my worksheet of Classical Mechanics, it must be elementary. I, however, am stunned and do not have the slightest idea of approaching this problem.

• This is certainly not a chaotic system. Is this part one of a question that goes on to have more parts? Nov 30, 2014 at 21:18
• @garyp No, this is an independent question. Nov 30, 2014 at 21:19
• @garyp Can you please tell how do you figure out whether it is chaotic or not? Nov 30, 2014 at 21:22
• @Samurai just checking, this isn't a Lagrangian mechanics question is it?
– user12029
Nov 30, 2014 at 23:18
• @NeuroFuzzy Well, I haven't been taught that yet. So, I guess its not. Nov 30, 2014 at 23:41

First, look at the following Free Body Diagram.

For the disk

$$\text{T}\text{R}=\text{I}_{\text{cm}}\alpha$$ where $\text{T}$ is the tension in the string, $\alpha$ is the angular acceleration in the disk and $\text{I}_{\text{cm}}$ is the moment of inertia of the disk about it's center, which equals $\frac{1}{2}\times(\text{2m})\times\text{R}^{2}$.

and,

$$\text{T}=(\text{2m})\text{a}_{\text{cm}}$$

where $\text{a}_{\text{cm}}$ is the acceleration of the center of mass of the disk.

Also, the end of the string which is attached to the disk will have an acceleration, say $\text{A}$, i.e, $$\text{A}=(2\text{m})\text{a}_{\text{cm}} +\alpha \text{R}$$

For the particle

Since Impulse equals change in momentum, the particle's velocity is given by

$$\text{v}=\frac{\text{J}}{\text{m}}=\sqrt{10}\text{m}\text{s}^{-1}$$

In the reference frame of the disk, the particle will try to move in a circular path at the initial instant, thus for the particle we have,

$$\text{T}+\text{m}\text{A}=\frac{\text{m}\text{v}^{2}}{(2\text{R})}$$

where $\text{m}\text{A}$ is the pseudo force.

Solving for $\text{a}_{\text{cm}}$ from the above equations, we get

$$\boxed{\text{a}_{\text{cm}}=4\text{m}\text{s}^{-2}}$$

• Amazing! This is the correct answer as given on the back of my worksheet. Thanks a ton!! Dec 1, 2014 at 0:16