applied maths LC 2006 paper

niallofthenine

hi guys, problem here, question 8b part i,i understand that the principle is total potential and total kinetic at each point is equal but in the marking scheme given here http://www.examinations.ie/archive/markingschemes/2006/LC020ALPO00EV.pdf says we should use h= 0.6 - 0.6cos60 however i based my reference point as the lowest point of the rod and therefor got h=0.9. this gives me a different angular velocity, is my method still correct? would i still get the marks? thanks niall

Robbiethe3rd

the total kinetic and potntial at each point is definitely not equal!

PE1 + KE1 = PE2 + KE2

If thats what you mean; just thought I shhould clarify!

Precision

Although the rod is 1.2 meters long, its centre-of-gravity is at its mid-point.

Since gravity acts in the vertical direction, the value h is the distance in the vertical direction between the location of the centre-of-gravity when the potential energy is at its maximum and when the potential energy is at its minimum.

The potential energy is at its maximum when the rod is at 60 degrees to the vertical.

The potential energy is at its minimum when the rod is vertical.


When the rod is vertical, the centre-of-gravity is 0.6 meters below the pivot in the vertical direction.

When the rod is at 60 degrees to the vertical, then the centre-of-gravity is 0.6cos60 meters below the pivot in the vertical direction.

Therefore the value of h, as defined above is 0.6 - 0.6cos60 meters.

h = 0.6 - 0.6cos60 meters

which is

h = 0.3 meters