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wait what was the circular motion question ?????

(edited 10 months ago)

Original post by bluewhale87

wait what was the circular motion question ?????

The one with the string and the pendulum I think

Original post by bluewhale87

wait what was the circular motion question ?????

The string and the particle one where you had to find the maximum tension.

Original post by ed729781

messed up circular motion (question 6) and i honestly don’t know about question 8? everyone i’ve spoken to seems to have gotten a different highest velocity

i think for the most part the paper was really good though considering i struggle w mechanics generally

i think for the most part the paper was really good though considering i struggle w mechanics generally

I think I got around 5.3 for max velocity of the block on q8 by considering: Work done by pulling force - Work done against friction - Elastic potential = Kinetic.

For the tension question I got T=mgsec(theta) then differentiated and showed max tension is at theta = 0 and I also got max velocity of the pendulum as something around 2 by doing mg(change in h)=1/2mv^2

Original post by raducorha

I think I got around 5.3 for max velocity of the block on q8 by considering: Work done by pulling force - Work done against friction - Elastic potential = Kinetic.

For the tension question I got T=mgsec(theta) then differentiated and showed max tension is at theta = 0 and I also got max velocity of the pendulum as something around 2 by doing mg(change in h)=1/2mv^2

For the tension question I got T=mgsec(theta) then differentiated and showed max tension is at theta = 0 and I also got max velocity of the pendulum as something around 2 by doing mg(change in h)=1/2mv^2

I got 5.3 and 2 but the 5.3 is wrong apparently?

i got 3.4? i assumed it was at the point when the string became taught

Original post by bluewhale87

i got 3.4? i assumed it was at the point when the string became taught

I think it’s at the equilibrium point

Original post by Fort21

I think it’s at the equilibrium point

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

Original post by bluewhale87

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

I dont think so cause there is a constant pulling force. There’s always a resultant force to the right before the equilibrium, and at equilibrium the acceleration is 0 due to F=ma, not velocity.

Original post by bluewhale87

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

Acceleration is 0 at the equilibrium point, so v is a maximum.

Original post by bluewhale87

i got 3.4? i assumed it was at the point when the string became taught

do you mean when it became slack? so when T=0

i did that and got 3.4 too

Original post by ed729781

how did you guys find a distance it moves to put into your energy equation if you didn’t use the tension from the string to the left of the block is 0

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

Original post by raducorha

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

Same

Original post by raducorha

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

i mean for the second part where you’re trying to find v not the show that it’s 3.9 meters

Original post by ed729781

i mean for the second part where you’re trying to find v not the show that it’s 3.9 meters

oh did you use the extension that you got when finding 3.9m (2.9) when finding v for the next question?

(edited 10 months ago)

Original post by ed729781

oh did you use the extension that you got when finding 3.9m (2.9) when finding v for the next question?

Yeah when calculating the elastic potential energy at the equilibrium

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