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# M2 help, vertical circles watch

1. Hi, can anyone help, I keep getting the wrong answer on this question and it is really annoying me. Here is a picture of the diagram:

"A Marble of mass 0.01kg is on top of a smooth hemispherical bowl and given an initial speed of 1 m/s. The bowl has a radius of 0.5m and centre o.
a) The marble descends 0.1m, find the speed. got this right and figured out that v^2=1+0.2g

b) "Find the normal reaction of the bowl at this point"

So far I resolved vertically using F=ma, getting: R-0.01g X cos (theta) = m (v^2/R)
Problem is I do not know what theta is
Attached Images

2. Oh sorry my bad. If you know the speed, it is tangent to the semi-circle at that point, draw a triangle and you can find theta
3. OMG. AT Point M u have a triangle. The hypotenuse is the radius, and you know the height it falls down is 0.1 m . Therefore the height is 0.4m from the ground. Can you solve Theta now ?
OMG. AT Point M u have a triangle. The hypotenuse is the radius, and you know the height it falls down is 0.1 m . Therefore the height is 0.4m from the ground. Can you solve Theta now ?
Theta= cos^-1(0.4/0.5) ?

The answer is supposed to be 0.0192N, when i sub that value of theta in I get 0.1376N, no where near right

R=0.01x(1+0.2g/0.5) +0.01g x cos theta
5. (Original post by daftndirekt)
Theta= tan^-1(0.4/0.5) ?
Are you sure its tan ? Doesnt seem like it to me
Are you sure its tan ? Doesnt seem like it to me
Sorry cos rather (I've written cos down on my page oddly :P ), It is still wrong,
cos theta= 0.4/ 0.5, meaning theta =38.86...

Resulting in R=0.1376. But it doesn't.
7. (Original post by daftndirekt)
Sorry cos rather (I've written cos down on my page oddly :P ), It is still wrong,
cos theta= 0.4/ 0.5, meaning theta =38.86...

Resulting in R=0.1376. But it doesn't.
10/10 for the arithmetic, but unfortunately your starting equation is incorrect.

As the mass is moving in the arc of a circle, the resultant force on it must be mv^2/r towards the centre.

So,

or re-arranging:

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Updated: April 5, 2011
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