ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below
ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below
Exactly what I did. Any idea how many marks I'd lose?
Exactly what I did. Any idea how many marks I'd lose?
Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root2 instead of root10/2
Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2
Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2
there wasn't an alpha in the question but you had to realise that the masses of the two wires didn't lie on the same line so they made different angles to the downward vertical (like they usually do for M5 pendulum questions - they're normally straight objects) because the wire was kinked. Normally in M5 questions - it's just a normal straight pendulum so the angle both masses make with the downward vertical are the same because they both lie on a straight rod or whatever, however here they aren't.
The centre of mass of the object lies on 1/2,1/2 not on the line joining the 2 masses. use sum of moments=moments of sum using B as the origin
m(0,a)+m(a,0)=2m(x,y) (x,y)=1/2a,1/2a----> this does not lie on the line joining the two masses on your diagram.
Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing
Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing
Yes you're right it does lie on the line joining the two masses. The distance should be root10/2 a however. Im curious as to how people got root 2 tho?
They seem to pretty much match mine but I'm pretty sure that the answer to 1 was -6,12.5 as it was (-1,5)+(5/2)*(-2,3)=(-6,12.5) (going from memory)
yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.
yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r.
Yes I gave them the velocity equation. I cannot remember what it was but it fitted the boundary conditions and when differentiated and subbed back in, gave the required value so I'm confident with it
yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.
Having checked the answer you agreed with for Q5, I do disagree. The centre of mass was a distance of root(10)/2 from the pivot. m(2a)+m(a)=2mx leads to x=1.5a then the centre of mass was 1.5a below L and by symmetry, 0.5a from l which gives root(10)/2.