Rod or object resting on curved surface

4 \includegraphics[max width=\textwidth, alt={}, center]{15ed1dfc-8188-4e20-9c0b-ce31af35f0b6-3_512_983_267_580} A uniform sphere rests on a horizontal plane. The sphere has centre \(O\), radius 0.6 m and weight 36 N . A uniform rod \(A B\), of weight 14 N and length 1 m , rests with \(A\) in contact with the plane and \(B\) in contact with the sphere at the end of a horizontal diameter. The point of contact of the sphere with the plane is \(C\), and \(A , B , C\) and \(O\) lie in the same vertical plane (see diagram). The contacts at \(A , B\) and \(C\) are rough and the system is in equilibrium. By taking moments about \(C\) for the system, show that the magnitude of the normal contact force at \(A\) is 10 N . Show that the magnitudes of the frictional forces at \(A , B\) and \(C\) are equal. The coefficients of friction at \(A , B\) and \(C\) are all equal to \(\mu\). Find the smallest possible value of \(\mu\).

5 \includegraphics[max width=\textwidth, alt={}, center]{34dd6523-7c0c-4842-bbda-56ad8d3f9766-10_456_684_264_731} A uniform \(\operatorname { rod } A B\) of length \(2 x\) and weight \(W\) rests on the smooth rim of a fixed hemispherical bowl of radius \(a\). The end \(B\) of the rod is in contact with the rough inner surface of the bowl. The coefficient of friction between the rod and the bowl at \(B\) is \(\frac { 1 } { 3 }\). A particle of weight \(\frac { 1 } { 4 } W\) is attached to the end \(A\) of the rod. The end \(B\) is about to slip upwards when \(A B\) is inclined at an angle \(\theta\) to the horizontal, where \(\tan \theta = \frac { 3 } { 4 }\) (see diagram).

1. By resolving parallel to the rod, show that the normal component of the reaction of the bowl on the rod at \(B\) is \(\frac { 3 } { 4 } W\).
2. Find, in terms of \(W\), the reaction between the rod and the smooth rim of the bowl.
3. Find \(x\) in terms of \(a\).

5 \includegraphics[max width=\textwidth, alt={}, center]{38694ab3-44cd-48d1-922a-d5eb09b62826-3_650_698_248_721} Two parallel vertical smooth walls \(E F\) and \(C D\) meet a horizontal plane at \(E\) and \(C\) respectively. A uniform smooth rod \(A B\), of weight \(2 W\) and length \(3 a\), is freely hinged to the horizontal plane at the point \(A\), between \(E\) and \(C\). The end \(B\) rests against \(C D\). A uniform smooth circular disc of weight \(W\) is in contact with the wall \(E F\) at the point \(P\) and with the rod at the point \(Q\). It is given that angle \(B A C\) is \(60 ^ { \circ }\) and that \(A Q = a\) (see diagram). The rod and the disc are in equilibrium in the same vertical plane, which is perpendicular to both walls. Show that

1. the magnitude of the reaction at \(P\) is \(\sqrt { } 3 W\),
2. the magnitude of the reaction at \(B\) is \(\frac { 7 \sqrt { } 3 } { 9 } W\). Find, in the form \(k W\), the magnitude of the reaction on \(A B\) at \(A\), giving the value of \(k\) correct to 3 significant figures.

9 A smooth hollow hemisphere, of radius \(a\) and centre \(O\), is fixed so that its rim is in a horizontal plane. A smooth uniform \(\operatorname { rod } A B\), of mass \(m\), is in equilibrium, with one end \(A\) resting on the inside of the hemisphere and the point \(C\) on the rod being in contact with the rim of the hemisphere. The rod, of length \(l\), is inclined at an angle \(\theta\) to the horizontal, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{85514b55-3f13-4746-a3ef-747239b64cca-6_453_828_559_591}

1. Explain why the reaction between the rod and the hemisphere at point \(A\) acts through \(O\).
2. Draw a diagram to show the forces acting on the rod.
3. Show that \(l = \frac { 4 a \cos 2 \theta } { \cos \theta }\).

5 A ship \(S\) is travelling with constant speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a course with bearing \(345 ^ { \circ }\). A patrol boat \(B\) spots the ship \(S\) when \(S\) is 2400 m from \(B\) on a bearing of \(050 ^ { \circ }\). The boat \(B\) sets off in pursuit, travelling with constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a straight line.

1. Given that \(v = 16\), find the bearing of the course which \(B\) should take in order to intercept \(S\), and the time taken to make the interception.
2. Given instead that \(v = 10\), find the bearing of the course which \(B\) should take in order to get as close as possible to \(S\). \includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-4_337_954_278_544} A uniform rod \(A B\) has mass \(m\) and length \(2 a\). The point \(P\) on the rod is such that \(A P = \frac { 2 } { 3 } a\). The rod is placed in a horizontal position perpendicular to the edge of a rough horizontal table, with \(A P\) in contact with the table and \(P B\) overhanging the edge. The rod is released from rest in this position. When it has rotated through an angle \(\theta\), and no slipping has occurred at \(P\), the normal reaction acting on the rod at \(P\) is \(R\) and the frictional force is \(F\) (see diagram).
3. Show that the angular acceleration of the rod is \(\frac { 3 g \cos \theta } { 4 a }\).
4. Find the angular speed of the rod, in terms of \(a , g\) and \(\theta\).
5. Find \(F\) and \(R\) in terms of \(m , g\) and \(\theta\).
6. Given that the coefficient of friction between the rod and the edge of the table is \(\mu\), show that the rod is on the point of slipping at \(P\) when \(\tan \theta = \frac { 1 } { 2 } \mu\). \includegraphics[max width=\textwidth, alt={}, center]{181fad74-6e60-4435-a176-3edff5062c32-5_677_624_269_753} A smooth circular wire, with centre \(O\) and radius \(a\), is fixed in a vertical plane. The highest point on the wire is \(A\) and the lowest point on the wire is \(B\). A small ring \(R\) of mass \(m\) moves freely along the wire. A light elastic string, with natural length \(a\) and modulus of elasticity \(\frac { 1 } { 2 } m g\), has one end attached to \(A\) and the other end attached to \(R\). The string \(A R\) makes an angle \(\theta\) (measured anticlockwise) with the downward vertical, so that \(O R\) makes an angle \(2 \theta\) with the downward vertical (see diagram). You may assume that the string does not become slack.
7. Taking \(A\) as the level for zero gravitational potential energy, show that the total potential energy \(V\) of the system is given by $$V = m g a \left( \frac { 1 } { 4 } - \cos \theta - \cos ^ { 2 } \theta \right) .$$
8. Show that \(\theta = 0\) is the only position of equilibrium.
9. By differentiating the energy equation with respect to time \(t\), show that $$\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \frac { g } { 4 a } \sin \theta ( 1 + 2 \cos \theta ) .$$
10. Deduce the approximate period of small oscillations about the equilibrium position \(\theta = 0\).