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[muskan]

69 and 70. I tried it but i am not getting the answer. 

69

Edited September 26, 2018 by muskan

[swansont]

For 69 your energy balance is wrong. Why do you have -mgh? What is the speed at maximum height? 

You also don't show where the cos term comes from.

[studiot]

For 70 attached is a plan of action.

The area of interes forms a curved triangle which I have labelled ABD.
The centroid of ABD is labelled T which has an x coordinate  [math]{\overline X _T}[/math]

The area under the graph of   [math]y = \frac{{{x^2}}}{2}[/math]   forms another curved troangle I have labelled DBC.
This has centroid V with x coordinate

Note that these two curved triangles fit together to form a 2 x 2 square (I have labelled this ABCD)
This square has centroid S with standard values for its centroidal position.

 

Now the trick is to realise that the moment of this square must equal the sum of the moments of its two componeent curved triangles.

This should be a good start for you.

 

 

As to question 69, Are you sure you have presented the problem correctly?

The half metre length of string must go slack at the intial throw and only become tensioned when the mass has fallen back.
The questions states that theta is measured to the left, ie when the mass is hanging on 1.5 m of string and swings to the left, clear of the peg.

This would make the radius of circular swing 1.5 m.

Edited September 26, 2018 by studiot

[muskan]

In 70, the the plate is defined by x=0, y=2 and y= 1/2 x^2. Sir,  you took y=0 instead of x=0.

[studiot]

5 hours ago, muskan said:

In 70, the the plate is defined by x=0, y=2 and y= 1/2 x^2. Sir,  you took y=0 instead of x=0.

 

Look carefully

AD is the y axis or the line x = 0.

AB is the line y = 2

DB is the line y = 1/2X²

I said this was the area of interest, ie the plate.

16 hours ago, studiot said:

The area of interes forms a curved triangle which I have labelled ABD.

 

[studiot]

21 hours ago, studiot said:

As to question 69, Are you sure you have presented the problem correctly?

The half metre length of string must go slack at the intial throw and only become tensioned when the mass has fallen back.
The questions states that theta is measured to the left, ie when the mass is hanging on 1.5 m of string and swings to the left, clear of the peg.

This would make the radius of circular swing 1.5 m.

Actually this question is simpler than I first thought, but your diagram is all wrong as I said.

However physical reasoning about energy is definitely the best way to solve this.

I make the answer to be A,

[math]\cos \theta  = \frac{1}{6}[/math]

No complicated maths is needed.

Edited September 27, 2018 by studiot