The diagram shows a section of a curtain track in a vertical plane. The curved section, CDE, forms a circular arc of radius of curvature 0.75 m and the point D is 0.25 m higher than B. A ball-bearing of mass 0.060 kg is released from A, which is 0.50 m higher than B. Assume that rotational and frictional effects can be ignored and that the ball-bearing remains in contact with the track throughout the motion.
(a) Calculate the speed of the ball-bearing (i) at B, (ii) at D.
(b) Draw a diagram showing the forces acting on the ball-bearing when it is at D and calculate the reaction between the track and the ball-bearing at this point.
Answers: (a) (i) 3.2 m s-1, (ii) 2.2 m s-1, (b) 0.2 N
2. The attempt at a solution
For (a) I used the following formula: v2 = u2 + 2 a s
(i) u / v0 = 0, a = 10 m s-2 (positive, since the ball is moving downwards), s = hA = 0.5 m
Plug in and get: v2 = 0 + 2 * 10 * 0.5 = 10 → v = 3.2 m s-1
(ii) u = 3.2 m s-1, a = -10 m s-2 (since the ball is moving upwards), s = h D = 0.25 m
v2 = 3.22 + 2 * (-10) * 0.25 = 5 → v = 2.2 m s-1
I think the (a) part should be correct.
In terms of (b) I am not sure. I used the F = m v2 / r formula. m = 0.060 kg, v = 2.2 m s-1, r = 0.75 m.
Plug in and get: F = (0.060 * 2.22) / 0.75 = 0.4 N, however, the answer is 0.2 N.
And this is how I see the forces acting on the ball, however, again, not sure.
I have the weight (mg) and the normal reaction as the vertical forces. But regarding the horizontal ones I've got two question marks.
Any help please?