Can anyone please help with the attatched question?
Thanks for any help you can give.
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M2 OCR question: Circular motion in a hollow sphere watch
- Thread Starter
- 21-06-2005 23:17
- 21-06-2005 23:59
The radius of the circle is rt(5²-3²) = 4m. Take t to be the angle the normal makes with the radius.
R sint = mg
R (3/5) = 0.5g
=> R = (5/6)g N
R cost = mv²/r
(5/6)g (4/5) = 0.5v²/4
=> v =~ 7.23 m/s
Since v=rw, then:
w = v/r = 7.23/4 = 1.81 rad/s
T = 2pi/w = 2pi/1.81 = 3.47 s
- 22-06-2005 01:41
(i) this is not correct - sorry i havnt got the formula infront of me, but you just worked out the component of gravity that is acting onto the side of the sphere. you also need to take into account the centripetal force exerted on the particle by the bowl.
so you need to work out the force of the particle pushing outward, and then find the component of this at normal to the edge of the bowl. you need to add this to the component of gravity.
correct me if im wrong please, im not completely certain.
- 22-06-2005 03:17
I didn't work out the component of gravity. I simply used it to work out the normal reaction exerted by the sphere.