M3 June 2014 Q1
1.
A hemispherical bowl of internal radius \(4r\) is fixed with its circular rim horizontal. The centre of the circular rim is \(O\) and the point \(A\) on the surface of the bowl is vertically below \(O\). A particle \(P\) moves in a horizontal circle, with centre \(C\), on the smooth inner surface of the bowl. The particle moves with constant angular speed \(\sqrt{\dfrac{3g}{8r}}\)
The point \(C\) lies on \(OA\), as shown in Figure 1.
Find, in terms of \(r\), the distance \(OC\). (9)
| Scheme | Marks |
|---|---|
| \(R\sin\theta = m \times 4r\sin\theta \times \dfrac{3g}{8r}\) | M1A1A1 |
| \(R = \dfrac{3}{2}mg\) | |
| \(R\cos\theta = mg\) | M1A1 |
| \(\dfrac{3}{2}mg\cos\theta = mg\) | M1(dep) |
| \(\cos\theta = \dfrac{2}{3}\) | A1 |
| \(OC = 4r\cos\theta = 4r \times \dfrac{2}{3} = \dfrac{8}{3}r\) oe | M1A1 |
| (9 marks) |
Notes
M1 for NL2 towards \(C\) - Accept use of \(v = \sqrt{\dfrac{3g}{8r}}\) and \(a = \dfrac{v^2}{r}\) as a mis-read
A1 for LHS fully correct
A1 for RHS fully correct
ALT: Work in the direction of \(R\) and obtain the same equation with \(\sin\theta\) "cancelled". Give M1A1A1 if fully correct, M0 otherwise.
M1 for resolving vertically
A1 for the equation fully correct
M1 dep for eliminating \(R\) between the two equations Dependent on both above M marks
A1 for \(\cos\theta = \dfrac{2}{3}\)
M1 for attempting to use trig or Pythagoras to obtain \(OC\)
A1 cso for \(OC = \dfrac{8}{3}r\)
Alternative for Question 1
| \(R\sin\theta = m \times a \times \dfrac{3g}{8r}\) | M1A1A1 |
| \(R\cos\theta = mg\) | M1 A1 |
| \(\tan\theta = \dfrac{3a}{8r}\) | M1 A1 |
| \(\dfrac{a}{OC} = \dfrac{3a}{8r}\) | M1 |
| \(OC = \dfrac{8r}{3}\) | A1 |