A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is \(20\text{ m}\). The speed of the car at the top of the hill is between:
(2008)
1. \(13\text{ m/s}\) and \(14\text{ m/s}\)
2. \(14\text{ m/s}\) and \(15\text{ m/s}\)
3. \(15\text{ m/s}) and \(16\text{ m/s}\)
4. \(16\text{ m/s}\) and \(17\text{ m/s}\)
View Answer
At the top of a hill, "weightlessness" implies the normal force is zero. The centripetal force is provided solely by gravity: (mg = \frac{mv^2}{R}). Thus, (v = \sqrt{gR}). Using (g = 9.8\text{ m/s}^2) and (R = 20\text{ m}), (v = \sqrt{9.8 \times 20} = \sqrt{196} = 14\text{ m/s}). This speed lies between \(14\text{ m/s}\) and \(15\text{ m/s}\).
A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
(2019)
1. The mass is at the highest point
2. The wire is horizontal
3. The mass is at the lowest point
4. Inclined at an angle of $60^\circ$ from vertical
View Answer
Tension in the string/wire is maximum at the lowest point of the vertical circle ($T = mg + \frac{mv^2}{r}$), making it most likely to break there.
A stone is tied to a string of length ‘$\ell$’ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed ‘$u$’. The magnitude of the change in velocity as it reaches a position where the string is horizontal ($g$ being acceleration due to gravity) is:
(2004)
1. $\sqrt{u^2 - g\ell}$
2. $u - \sqrt{u^2 - 2g\ell}$
3. $\sqrt{2g\ell}$
4. $\sqrt{2(u^2 - g\ell)}$
View Answer
Velocity at horizontal position is $v_2 = \sqrt{u^2 - 2g\ell}$. Since velocities are perpendicular, magnitude of change in velocity is $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.