Standard +0.3 This is a standard vertical circle problem requiring energy conservation and the condition for complete revolution (minimum speed at top). Students must apply conservation of energy between the release point and the lowest speed point (top of circle), using the given constraint. The setup is straightforward with clear given values, requiring only routine application of well-practiced techniques without novel insight.
3 A smooth wire is shaped into a circle of radius 4.2 m which is fixed in a vertical plane with its centre at a point \(O\). A small bead \(B\) is threaded onto the wire. \(B\) is held so that \(O B\) makes an angle of \(\frac { 1 } { 3 } \pi\) radians with the downwards vertical through \(O\).
\(B\) is projected downwards along the wire with initial speed \(u \mathrm {~ms} ^ { - 1 }\) (see diagram). In its subsequent motion \(B\) describes complete circles about \(O\).
\includegraphics[max width=\textwidth, alt={}, center]{98053e88-1aec-4b0d-ae5f-ece4ad340266-3_493_665_561_242}
Given that the lowest speed of \(B\) in its motion is \(4 \mathrm {~ms} ^ { - 1 }\) determine the value of \(u\).
Or \(\frac{1}{2}mu^2 = \frac{1}{2}m\times4^2 + m\times9.8\times4.2\times(1+\cos\frac{\pi}{3})\); equating their gain of PE with their loss of KE (signs must be correct)
\(u > 0 \Rightarrow u =\) awrt \(11.8\)
A1 [5]
# Question 3:
Initial PE $= m\times9.8\times4.2(1-\cos\pi/3)$ | M1 (AO 3.1b) | Calculation of initial energy; assuming that the lowest point is the 0 PE level; do not allow use of suvat
Speed is lowest when $B$ reaches the top | B1 (AO 2.2a) | oe soi e.g. cons of energy seen; if not stated explicitly, then award for any energy equation that leads to $u > 8$
Energy at top $= \frac{1}{2}\times m\times4^2 + m\times9.8\times(2\times4.2)$ | M1 (AO 1.1) | $(= 90.32m)$; adding PE and KE at the top
$m\times9.8\times4.2(1-\cos\pi/3) + \frac{1}{2}mu^2 =$ their energy at top | M1 (AO 1.1) | $(20.58m + \frac{1}{2}mu^2 = 90.32m)$ adding PE and KE at start and equating; consistent dimensions
$u > 0 \Rightarrow u =$ awrt $11.8$ | A1 [5] (AO 1.1) | Must be positive; $u^2 = 139.48$
**Alternative solution:**
Change in PE $= m \times 9.8 \times 4.2 \times (1 + \cos\frac{\pi}{3})$ | M1 | $61.74m$; i.e. initial position has zero GPE
Speed is lowest when $B$ reaches the top | B1 |
Change in KE $= \pm\frac{1}{2}m(4^2 - u^2)$ | M1 | May be seen in balanced equation
$m\times9.8\times4.2\times(1+\cos\frac{\pi}{3}) = -\frac{1}{2}m(4^2 - u^2)$ oe | M1 | Or $\frac{1}{2}mu^2 = \frac{1}{2}m\times4^2 + m\times9.8\times4.2\times(1+\cos\frac{\pi}{3})$; equating their gain of PE with their loss of KE (signs must be correct)
$u > 0 \Rightarrow u =$ awrt $11.8$ | A1 [5] |
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3 A smooth wire is shaped into a circle of radius 4.2 m which is fixed in a vertical plane with its centre at a point $O$. A small bead $B$ is threaded onto the wire. $B$ is held so that $O B$ makes an angle of $\frac { 1 } { 3 } \pi$ radians with the downwards vertical through $O$.\\
$B$ is projected downwards along the wire with initial speed $u \mathrm {~ms} ^ { - 1 }$ (see diagram). In its subsequent motion $B$ describes complete circles about $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{98053e88-1aec-4b0d-ae5f-ece4ad340266-3_493_665_561_242}
Given that the lowest speed of $B$ in its motion is $4 \mathrm {~ms} ^ { - 1 }$ determine the value of $u$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2022 Q3 [5]}}