Vertical Circular Motion (5 pages, 14/9/2013)
(1) Particle on a string
An equivalent situation is where the particle is inside a circular track. In
both cases, the force on the particle provided by the string/track can only
be towards the centre of the circle. In the case of the string the force is the
tension in the string; in the case of the track it is the normal reaction on the
particle (the track can push against the particle, but not pull it).
We shall assume that the particle starts at the bottom of the circle with
speed u, and that the circle has radius r.
Possible scenarios
(a) the particle comes to a halt and falls back to oscillate on an arc of the
circle
(b) the string becomes slack and the particle falls inwards
(c) the particle reaches the top of the circle, in which case it continues its
circular motion indefinitely (ignoring resistances to motion)
Condition for the particle to reach a horizontal position
Let v be the speed for a given θ. Clearly v decreases as θ increases from 0
to π, and then increases as θ increases from π to 2π .
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Referring to diagram (a) above, the radial equation of motion is:
F – mg cos θ = mv² / r (A)
and Conservation of Energy gives:
½ mu² - mgr = ½ mv² - mgrcos θ
(taking the centre of the circle as the zero of PE)
⇒ v² = u² - 2gr(1 – cos θ) (B)
Note that Conservation of Energy is often used to solve problems
concerning circular motion where the tension in the string etc is not
needed (as with other topics in Mechanics, it can turn a difficult-looking
problem into something simple).
When θ ≤ π/2, cos θ ≥ 0 and (A) ⇒ F ≥ mv² / r
Hence, F=0 will not occur before v=0, and the string will not become slack
before the particle comes to a halt (and then falls back along its original
path); ie the condition to investigate is whether v > 0.
From (B), v > 0 ⇒ u² > 2gr(1 – cos θ)
This is true for all θ ≤ π/2, provided that u² > 2gr (since 1-cos θ has its
largest value [of 1] at θ = π/2).
This is the condition for the particle to reach a horizontal position.
Condition for the particle to reach the top of the circle
When π /2 < θ ≤ π (see diagram (b) above):
First of all, note that
F + mgcos  = F + mgcos( π -θ) = F – mgcos θ
so that equation (A) still applies.
For π /2 < θ ≤ π, cos θ < 0 and (A) ⇒ F < mv² / r. Hence F = 0 will occur
before v=0; ie the string will become slack before the particle can come to
a halt, and the condition to investigate is whether F > 0.
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When F>0, (A) & (B) ⇒ u² > gr(2 – 3cos θ)
Proof [this could be skipped]
(A) ⇒ mv² / r + mg cos θ > 0
Hence, substituting for v² from (B),
⇒
m 2
(u  2 gr (1  cos  ) ) + mg cos θ > 0
r
m 2
(u  2 gr  3gr cos  ) > 0
r
⇒ u 2  gr (2  3 cos )  0
and 𝑢2 > gr(2 – 3cos θ)
Then, in order that F > 0 for all π /2 < θ ≤ π, u² > 5gr [ 2 – 3cos θ has its
largest value (of 5) at θ = π ]
This is the condition for the particle to reach the top.
Alternative methods of investigating whether the particle reaches the top
of the circle:
(a) Determine whether F > 0 at θ = π
(b) Determine when F = 0 (if at all)
(Strictly speaking though, both of these methods make the assumption that
the particle hasn’t come to a halt before reaching the top.)
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(2) Particle threaded on a wire
This is equivalent to being attached to the end of a light rod.
Here the reaction on the particle can be in either direction, provided the net
force is directed towards the centre.
The possible scenarios (from (1)) are now (a) and (c), and the only
condition to investigate is whether v > 0.
Condition for the particle to reach the top of the circle
When v > 0, (B) ⇒ u² > 2gr(1 – cos θ)
Hence, in order that v > 0 for all θ ≤ π, u² > 4gr
[ 1 – cos θ has its largest value (of 2) at θ = π ]
This is the condition for the particle to reach the top.
Alternative method of investigating whether the particle reaches the top of
the circle: Determine whether v > 0 at θ = π (if v < 0, this means that the
particle came to a halt before reaching the top; v of course cannot actually
be negative at the top).
(3) Particle on the outside of a circular track
Here the normal reaction on the particle is directed away from the centre
(the track cannot pull the particle inwards), so that F in the diagrams has to
be negative.
When θ ≤ π/2, it is clear that, for circular motion to occur, F must be
positive, otherwise the net force would be away from the centre and
circular motion would not be possible. Hence the particle cannot remain on
the outside of a circular track at these angles.
When F < 0, (A) ⇒ mv² / r + mg cos θ < 0 ⇒ v² < - grcos θ
As θ ranges from π /2 to π , the maximum allowable speed of
√(−grcosθ) ranges from 0 to √gr.
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Thus, if the particle starts on the track and continues upwards, then it will
stay on the track at least until it reaches the top, since its speed will be
falling (by Conservation of Energy, since the P.E. is increasing) and the
maximum allowable speed will be rising.
For θ > π :
To find the angle θ (still measured from the bottom of the circle) at which
the particle leaves the track, supposing that the particle has speed w at the
top:
This occurs when F=0.
(A) ⇒ mv² / r + mg cos θ = 0 ⇒ v² = -gr cos θ = 0 (C)
and Conservation of Energy gives:
½ mw² + mgr = ½ mv² - mgrcos θ (treating the centre of the circle as the
zero of PE), and noting that θ is measured from the bottom of the circle
⇒ v² = w² + 2gr(1 + cos θ ) (D)
(C) & (D) then ⇒ w² + 2gr(1 + cos θ) = -grcos θ
⇒ w² + 2gr = -3grcos θ
⇒ cos θ = - (w² + 2gr)/3gr
Special cases
(a) w = 0 ⇒ cos θ = -2/3 ⇒ θ = 228.2° (or 48.2° measured from the top)
(b) θ = 1800 ⇒ w² = gr
ie the particle will leave the track at the top if its speed is √gr ,
which is consistent with the maximum allowable speed found earlier.
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