Conical Pendulum Problem (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/ConicalPendulum1.jpg

I included diagram. Can someone do iii) and iv)

for iv) i have no idea what happens if the ring is free to move, someone care to explain?

 

[Yip]

My mechanics are a bit rusty, but i think this is right...
For part c, u kno that T1>0, but T2 may not necessarily be 0. Since both tensions have to be positive, set T2>0.
T2=mrw^2-mgsin(theta)>0
rw^2>gtan(theta)
Since r=4asin(theta),
4aw^2sin(theta)>gtan(theta)
4aw^2>gsec(theta)=5g/4
Therefore 16aw^2>5g

Part d is asking what condition does the new angular velocity v have to satisfy so that the unfastened ring remains at exactly the same spot as before. That will happen if both tensions equal each other, i.e. T1=T2

Setting T1=T2,
rv^2[cos(theta)-sin(theta)]=g[sin(theta)+cos(theta)]
4a(3/5)(1/5)v^2=g(7/5)
Therefore 12av^2=35g

 

[namburger]

yeppp, seems correct Yip

another problem:
A light inextensible string of length 3L is threaded through a smooth vertical ring which is free to turn. The string carries a particle at each end. One particle A of mass m is at rest at a distance L below the ring. The other particle B of mass M is rotating in a horizontal circle whose centre is A. Find the angular velocity of B and find m in terms of M.

I got the angular velocity of sqrt (g/L). Can someone help me with finding m in terms of M

Good practice ^^

 

[duy.le]

i get sqrt(mg/2ML), dont know if that is right.

 

[namburger]

WRONG DUY LE :O

bad luck, i won't tell you answer coz its never good to work towards the answer.

 

[duy.le]

ill do it later then.

 

[duy.le]

yeppp, seems correct Yip

another problem:
A light inextensible string of length 3L is threaded through a smooth vertical ring which is free to turn. The string carries a particle at each end. One particle A of mass m is at rest at a distance L below the ring. The other particle B of mass M is rotating in a horizontal circle whose centre is A. Find the angular velocity of B and find m in terms of M.

I got the angular velocity of sqrt (g/L). Can someone help me with finding m in terms of M

Good practice ^^

what does that mean... does that mean in the same plane or simply that the center of its rotation is "a" . i.e "a" but vertically down.