Look at the diagram properly. The reaction force does come out diagonally/perpendicular to the cylinder, as you've rightly said; but in their diagram, they've drawn two arrows resolving that perpendicular/diagonal force horizontally and vertically. I've zoomed it up for you:
As you can see, they've split the normal perpendicular force into the horizontal and the vertical component.
But isn't friction acting between the rod and the cylinder supposed to be (imagine in they didn't split the normal force) acting perpendicular to the normal?
But isn't friction acting between the rod and the cylinder supposed to be (imagine in they didn't split the normal force) acting perpendicular to the normal?
Yes, but they have also split the friction into horizontal and vertical, look at the first diagram of the model answers.
Have they made friction act in the direction which acts "into" the cylinder? (I find their diagram confusing, hence I'm here :/ )
The rod wants to slip off the cylinder in the obvious intuitive direction/way. Hence friction will oppose this and will act parallel to the rod (perpendicular to the normal, as you said up the rod.) When you resolve the parallel force up the rod, the force acts vertically upwards and horizontally rightwards (or into the cylinder). Sketch it for yourself and see.
As for the (d) of the projectiles question, the speed is the square root of the sum of the squares of the the horizontal and vertical speed. Your horizontal speed seems spot on! But your vertical speed is totally off.
So we have (resolving horizontally) a constant horizontal speed of 27agcosθ and a vertical speed (here we need to use SUVAT in the vertical direction:
v2=u2+2as=27agsin2θ−2g×6a=27agsin2θ−12ag.
Now we know that cosθ=261 so sketching a little triangle tells us that sinθ=265
Hence the resultant speed is 27ag(265)2−12ag+27ag261
The rod wants to slip off the cylinder in the obvious intuitive direction/way. Hence friction will oppose this and will act parallel to the rod (perpendicular to the normal, as you said up the rod.) When you resolve the parallel force up the rod, the force acts vertically upwards and horizontally rightwards (or into the cylinder). Sketch it for yourself and see.
Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
Sorry, I'm crap at this moments stuff! I'll let ghostwalker answer this one.
As for the (d) of the projectiles question, the speed is the square root of the sum of the squares of the the horizontal and vertical speed. Your horizontal speed seems spot on! But your vertical speed is totally off.
So we have (resolving horizontally) a constant horizontal speed of 27agcosθ and a vertical speed (here we need to use SUVAT in the vertical direction:
v2=u2+2as=27agsin2θ−2g×6a=27agsin2θ−12ag.
Now we know that cosθ=261 so sketching a little triangle tells us that sinθ=265
Hence the resultant speed is 27ag(265)2−12ag+27ag261
Ah yes, I've got the right answer now! Thanks Did you figure out the q8 part c I posted (it was in the same post as this speed question)
Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
If the line of action of the force acts along the rod, what's the perpendicular distance of that line of action from A?
Here's some examples. In each case the perpendicular distance is from the origin to the line of action (red), and is repressented by the line in green. You can see (last diagram) that when our line of action goes through the origin, the perpendicular distance is zero, and so the moment of that force is zero.
Here's some examples. In each case the perpendicular distance is from the origin to the line of action (red), and is repressented by the line in green. You can see (last diagram) that when our line of action goes through the origin, the perpendicular distance is zero, and so the moment of that force is zero.
I've got the answer now, thanks
Would you mind giving me hand on part B? I managed to get part a correct ?(3cm)