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R Gopalakrishnan Grade: 11
```        Sir, I was going through some questions in H.C.Verma and have a few doubts as follows:
1) A disk rotates about its axis with a constant angular acceleration of 4 rad/s^2.Find the radial and tangential accelerations of a particle at a distance of 1 cm from the axis at the end of the first second after the disk starts rotating.
I used the formula @=a/R
where:
@=tangential acceleration
a=angular acceleration
R=radius
Susbstituting the values i got the correct answer 4 cm/s^2. However how do you find the radial acceleration?
2)Three particles,each of mass 200g are kept at the corners of  an equilateral triangle of side 10cm.Find the moment of Inertia of the system of an axis
a)joining two of the particles and
b)passing through one of the particles and perpendicular to the plane of the particles.
For this question I drew the figure and tried to go about it but couldn't get it. The answers given in the book are:
a)1.5*10^-3 kg/m^2
b)4.0*10^-3  kg/m^2
3)Particles of masses 1g,2g,3g,.....,100g are kept at the maks 1cm,2cm,3cm....100cm respectively on a metre scale.Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.
The answer given is:0.43 kg/m^2
```
8 years ago

## Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
```										Dear R Gopalakrishna
Question 1
@=a/R  this is not correct fornula
@=aR  is the correct formula
and for radial acceleration
w=w0 +at  where w angular velocity at time t
and w0 at time t=0 sec
so w =0+4*1
=4 rad/sec
and the radial acceeraion =w2R
=16*1=16 cm/sec2
Please ask seperate question in seperate post

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.  All the best.  Regards, Askiitians Experts Badiuddin
```
8 years ago
Badiuddin askIITians.ismu Expert
147 Points
```										Dear gopal
second question

For case a
I =0.200 X (5√3 *10-2)2
=1.5 X 10-3
For case b
I= .2 X(10 *10-2)2     +.2X(10 *10-2)2
=4X10-3

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.  All the best.  Regards, Askiitians Experts Badiuddin

```
8 years ago
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