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RoryP
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Circular Motion!
A bead of mass m moves on a smooth circular ring of radius a which is fixed in a vertical plane. Its speed at A, the highest point of its path, is v and its speed at B, the lowest part of its path, is 7v.
a) show that v= [square root](ag/12)
b) find the reaction of the ring on the bead, in terms of m and g, when the bead is at A
F=mv 2 /r
EK= 1/2mv2
EP= mgh
a)
At point A
EK= 1/2mv2
EP= 2mga
At point B
GPE= 0
EK= 1/2m(7v)2
(Conservation of energy)
1/2m(7v)2 = 1/2mv2 + 2mga
(m's cancel)(multiply through by 2)
49v2= v2 + 4ag
48v2= 4ag
v2= 4ag/48 = ag/12
v=[square root](ag/12)
That I am fine with, its part b which confuses me!
b)
mv2/r + mg = R
(r=a)(v2= ag/12)
mag/12a + mg= R
mg/12 + mg= R
R=13mg/12
But that isn't correct! any help would be greatly appreciated =]
Homework Statement
A bead of mass m moves on a smooth circular ring of radius a which is fixed in a vertical plane. Its speed at A, the highest point of its path, is v and its speed at B, the lowest part of its path, is 7v.
a) show that v= [square root](ag/12)
b) find the reaction of the ring on the bead, in terms of m and g, when the bead is at A
Homework Equations
F=mv 2 /r
EK= 1/2mv2
EP= mgh
The Attempt at a Solution
a)
At point A
EK= 1/2mv2
EP= 2mga
At point B
GPE= 0
EK= 1/2m(7v)2
(Conservation of energy)
1/2m(7v)2 = 1/2mv2 + 2mga
(m's cancel)(multiply through by 2)
49v2= v2 + 4ag
48v2= 4ag
v2= 4ag/48 = ag/12
v=[square root](ag/12)
That I am fine with, its part b which confuses me!
b)
mv2/r + mg = R
(r=a)(v2= ag/12)
mag/12a + mg= R
mg/12 + mg= R
R=13mg/12
But that isn't correct! any help would be greatly appreciated =]
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