A2 October 2020 Q4
4.

A uniform solid cylinder of base radius \(r\) and height \(\dfrac{4}{3}r\) has the same density as a uniform solid hemisphere of radius \(r\). The plane face of the hemisphere is joined to a plane face of the cylinder to form the composite solid \(S\) shown in Figure 3. The point \(O\) is the centre of the plane face of \(S\).
The solid \(S\) is placed with its plane face on a rough horizontal plane. The coefficient of friction between \(S\) and the plane is \(\mu\). A horizontal force \(P\) is applied to the highest point of \(S\). The magnitude of \(P\) is gradually increased.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{4}{3}\pi r^3\times\dfrac{2}{3}r + \dfrac{2}{3}\pi r^3\times\left(\dfrac{4}{3}r + \dfrac{3}{8}r\right) = \left(\dfrac{4}{3} + \dfrac{2}{3}\right)\pi r^3\times d\) | M1 | 2.1 |
| \(\left(\dfrac{8}{9}r + \dfrac{8}{9}r + \dfrac{1}{4}r = 2d\right)\) \(\left(\dfrac{73}{36}r = 2d\right)\) | A1 A1 | 1.1b 1.1b |
| \(\Rightarrow d = \dfrac{73}{72}r\) * | A1* | 2.2a |
| (4) |
Notes
M1: Moments equation. Dimensionally correct.
A1: Unsimplified equation with at most one slip
A1: Correct unsimplified equation
A1*: Obtain given result from correct working.
| Scheme | Marks | AO |
|---|---|---|
![]() | ||
| Resolving: \(\leftrightarrow F = P\), \(\updownarrow R = Mg\), \(F_{\max} = \mu R = \mu Mg\) | M1 | 1.1b |
| Slides if \(P \gt \mu Mg\) | A1 | 1.2 |
| Moments: \(\dfrac{7}{3}rP = rMg \qquad\) Tilts if \(P \gt \dfrac{3}{7}Mg\) | B1 | 1.1b |
| Comparison of restrictions to determine values of \(\mu\) | M1 | 3.1a |
| Slides first if \(\mu Mg \lt \dfrac{3}{7}Mg,\quad (0 \lt)\,\mu \lt \dfrac{3}{7}\) | A1 | 2.2a |
| (5) | ||
| (9 marks) |
Notes
M1: Resolve and use \(F = \mu R\) to find values of \(P\) for sliding
A1: Use the model to form the correct inequality
B1: Correct inequality for tilting
M1: Correct comparison of when it tilts and when it slides
A1: Correct conclusion
