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# Help needed with circular motion question!!

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1. Help needed with circular motion question!!
I have no idea how to proceed with this and the only method I could think of doesn't yield the right answer. I tried using Newtons second law,

i.e. T - Fr = mrw2

find two equations for T and solve, but that answer is wrong..
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2. Re: Help needed with circular motion question!!
w=6
3. Re: Help needed with circular motion question!!
7 (i) 0.3ω^2× 0.5 = T + 0.36 × 0.3g
0.2ω^2× 0.5 = T – 0.36 × 0.2g
0.1ω^2 × 0.5 = 0.36 × 0.5g
ω = 6
T = 0.3 × 6^2 × 0.5 – 0.36 × 0.3 × 10
T = 4.32

That's what mark scheme says, BUT I don't get why T would add to the friction?
4. Re: Help needed with circular motion question!!
P and Q have different masses. The force required to keep a more massive object in circular motion is greater than that to keep a less massive one (F=mrw^2). The tension in the string is constant, so it acts equally on each particle.

Hence the lighter particle must be kept in the circular orbit by tension inwards and friction outwards, and the heavier one by tension and friction both inwards. (To get the greatest difference between the centripetal forces, and hence the greatest value of w)

The frictional force (As the normal force = weight = mg)

hence

insert value of r and to find w

For the second part - as there is no friction, the two masses must move in a circle around their combined centre of mass (like an orbit)
Alternatively you can find simultaneous equations to solve (they're easy ones)
Last edited by The Mr Z; 23-05-2012 at 19:55.
5. Re: Help needed with circular motion question!!
(Original post by The Mr Z)
P and Q have different masses. The force required to keep a more massive object in circular motion is greater than that to keep a less massive one (F=mrw^2). The tension in the string is constant, so it acts equally on each particle.
I understand this.
Hence the lighter particle must be kept in the circular orbit by tension alone, and the heavier one by tension and friction. (To get the greatest difference between the centripetal forces, and hence the greatest value of w)
Sorry I don't understand this, could you please elaborate??
The frictional force (As the normal force = weight = mg)

hence

insert value of r and to find w

For the second part - as there is no friction, the two masses must move in a circle around their combined centre of mass (like an orbit)
Alternatively you can find simultaneous equations to solve (they're easy ones)
Why would the lighter one have to be held by tension alone? There IS friction acting on it?
6. Re: Help needed with circular motion question!!
(Original post by bmqib)
Why would the lighter one have to be held by tension alone? There IS friction acting on it?
Wait yes, good point there is.

The lighter one would have to have friction acting outwards on it, and the heavier acting inwards, to give the maximum difference between the two centripetal forces. (we're looking for the maximum difference to give us the maximum value of w)

see above for correction
Last edited by The Mr Z; 23-05-2012 at 19:55.

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Last updated: May 23, 2012
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