M4 June 2014 Q7
7.
A bead \(B\) of mass \(m\) is threaded on a smooth circular wire of radius \(r\), which is fixed in a vertical plane. The centre of the circle is \(O\), and the highest point of the circle is \(A\). A light elastic string of natural length \(r\) and modulus of elasticity \(kmg\) has one end attached to the bead and the other end attached to \(A\). The angle between the string and the downward vertical is \(\theta\), and the extension in the string is \(x\), as shown in Figure 2.
Given that the string is taut,
Given also that \(k = 3\),
| Scheme | Marks |
|---|---|
| Measuring GPE from A, GPE \(= -mg\cos\theta(r + x)\) | B1 |
| EPE \(= \dfrac{kmgx^2}{2r}\) | B1 |
| From the isosceles triangle, \(\cos\theta = \dfrac{x + r}{2r}\) | B1 |
| \(V = -mg\cos\theta(r + x) + \dfrac{kmgx^2}{2r}\) | M1 |
| \(= -mg\cos\theta \times 2r\cos\theta + \dfrac{kmgr^2(2\cos\theta - 1)^2}{2r}\) | A1 |
| \(= mgr\left\{-2\cos^2\theta + 2k\cos^2\theta - 2k\cos\theta + \dfrac{k}{2}\right\}\) | |
| \(= 2mgr\{(k - 1)\cos^2\theta - k\cos\theta\} +\) constant ** | A1 |
| (6) |
Notes
B1 Or \(-2mgr\cos^2\theta\), or \(-mgr(1 + \cos 2\theta)\) or equivalent
M1 Correct unsimplified total
A1 In terms of \(r\) & \(\theta\)
A1 Reach given answer correctly
| Scheme | Marks |
|---|---|
| \(V = 2mgr\left(2\cos^2\theta - 3\cos\theta\right) +\) constant | |
| \(V' = 2mgr(-4\cos\theta\sin\theta + 3\sin\theta)\) | M1 A1 |
| \(V' = 0 \Rightarrow \sin\theta = 0\) or \(\cos\theta = \dfrac{3}{4}\) | M1 |
| \(\theta = 0\) or \(\theta = \pm 0.72\) rads | A3 |
| \(V'' = 2mgr(-4\cos 2\theta + 3\cos\theta)\) | M1 |
| \(\theta = 0,\ V'' = -2mgr < 0\), unstable equilibrium | A1 |
| \(\cos\theta = \dfrac{3}{4},\ \ V'' = \dfrac{7mgr}{2} > 0\), stable equilibrium | A1 |
| (9) | |
| (15 marks) |
Notes
M1 Differentiate \(V\)
M1 Derivative = 0 and solve for \(\theta\)
A3 -1 for each missing solution
M1 Second derivative of \(V\)
A1 Need to see \(-2mgr\) or equivalent
A1 Do not need to consider the symmetrical position as well