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# Mechanics M2 Circular Motion help watch

1. I can't seem to answer these questions. For 18. I've worked out a) as 19.7ms^-2 by using a=r(omega)^2 and that 1 revolution is pi rads^-1
I'm having trouble drawing the diagram as I'm not sure where the reaction force would go... any help?

With 19, I get a). You get a triangle with sides 0.1 and 0.5 and so tana=5. Once again, stuck with the rest...

2. (Original post by m2ishard)
I can't seem to answer these questions. For 18. I've worked out a) as 19.7ms^-2 by using a=r(omega)^2 and that 1 revolution is pi rads^-1
I'm having trouble drawing the diagram as I'm not sure where the reaction force would go... any help?

With 19, I get a). You get a triangle with sides 0.1 and 0.5 and so tana=5. Once again, stuck with the rest...
For the diagram, you'll have the person's weight, friction, and a reaction force (the centripetal force). Friction and weight must balance if the cylinder is vertical, and then the reaction force can be calulated using circular motion equation.

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