Circular motion — practice questions

Let the lowest point be A and the highest point be B. Let the mass be $m = 0.2$ kg and the radius be $r = 0.8$ m. The initial speed at A is $u = 7$ m/s.

(i) Find the speed at the highest point, B. We use the principle of conservation of energy. The gain in potential energy (PE) is equal to the loss in kinetic energy (KE). The vertical height gained from A to B is $2r = 2(0.8) = 1.6$ m.

Gain in PE = $mg(2r) = 0.2 \times 9.8 \times 1.6 = 3.136$ J. Loss in KE = $\frac{1}{2}mu^2 - \frac{1}{2}mv_B^2$, where $v_B$ is the speed at B.

$\frac{1}{2}m(u^2 - v_B^2) = mg(2r)$ $u^2 - v_B^2 = 2g(2r) = 4gr$ $7^2 - v_B^2 = 4 \times 9.8 \times 0.8 = 31.36$ $49 - v_B^2 = 31.36$ $v_B^2 = 49 - 31.36 = 17.64$ $v_B = \sqrt{17.64} = 4.2 \text{ m/s}$

(ii) Find the reaction force at the highest point, B. At the highest point B, two forces act on the bead: its weight ($mg$) acting downwards, and the normal reaction from the wire ($R$), also acting downwards (as the bead is on the inside of the wire). These two forces together provide the centripetal force, which is directed towards the centre (downwards).

Apply $F=ma$ towards the centre: $R + mg = \frac{mv_B^2}{r}$ $R = \frac{mv_B^2}{r} - mg$ $R = \frac{0.2 \times 17.64}{0.8} - 0.2 \times 9.8$ $R = \frac{3.528}{0.8} - 1.96$ $R = 4.41 - 1.96 = 2.45 \text{ N}$ Since R is positive, our assumed direction was correct. The reaction force is 2.45 N, acting downwards towards the centre of the circle.

Let $m = 1500$ kg, $r = 120$ m, and $\theta = 10°$. We need to find the speed $v$ where the friction force is zero.

Draw a force diagram. The forces acting on the car are its weight, $mg$, acting vertically downwards, and the normal reaction, $R$, acting perpendicular to the banked surface.

We resolve the forces into horizontal and vertical components. The centre of the circular path is in the horizontal plane.

Vertically (no acceleration): The upward component of R must balance the weight. $R \cos \theta = mg \quad (*)$

Horizontally (centripetal acceleration): The horizontal component of R provides the centripetal force. $R \sin \theta = \frac{mv^2}{r} \quad (**)$

To eliminate R and solve for v, we can divide equation (**) by equation (*): $\frac{R \sin \theta}{R \cos \theta} = \frac{mv^2/r}{mg}$ $\tan \theta = \frac{v^2}{gr}$

Now, rearrange and solve for v: $v^2 = gr \tan \theta$ $v^2 = 9.8 \times 120 \times \tan(10°)$ $v^2 = 1176 \times 0.1763...$ $v^2 \approx 207.35$ $v \approx \sqrt{207.35} \approx 14.4 \text{ m/s}$

The car can travel at approximately 14.4 m/s (to 3 s.f.) with no tendency to slip.