A tube of length \(L\) is filled completely with an incompressible liquid of mass \(M\) and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity \(omega\). The force exerted by the liquid at the other end is:
(2006)
1. \(\frac{ML\omega^2}{2}\)
2. \(\frac{ML\omega^2}{2}\)
3. \(2ML\omega^2\)
4. \(\frac{ML^2\omega^2}{2}\)
View Answer
Concept: Centrifugal force in a rotating system.
Formula: The force \(F\) is the integral of centrifugal force elements \(dF = dm \cdot r \cdot \omega^2\) from \(0\) to \(L\). Here, \(dm = (M/L)dr\).
Solution: \(F = \int_0^L \frac{M}{L} \omega^2 r dr = \frac{M\omega^2}{L} \left[\frac{r^2}{2}\right]_0^L = \frac{ML\omega^2}{2}\).
A stone tied to the end of a string of \(1 \text{ m}\) long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?
(2005)
1. \(\pi^2 \text{ m/s}^2\) and direction along the radius towards the centre
2. \(\pi^2 \text{ m/s}^2\) and direction along the radius away from the centre
3. \(\pi^2 \text{ m/s}^2\) and direction along the tangent to the circle
4. \(\pi^2 \text{ m/s}^2\) and direction along the radius towards the centre
View Answer
Concept: Centripetal acceleration in uniform circular motion.
Formula: \(a = \omega^2 r\), where \(\omega = 2\pi f\) and \(f\) is frequency.
Solution: Frequency \(f = 22 \text{ rev}/44 \text{ s} = 0.5 \text{ Hz}\). Angular velocity \(omega = 2\pi(0.5) = \pi \text{ rad/s}\). Radius \(r = 1 \text{ m}\). So, \(a = (\pi)^2 (1) = \pi^2 \text{ m/s}^2\). Direction is always towards the centre.
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 $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.
For a body angular velocity $\vec{\omega} = \hat{i} – 2\hat{j} + 3\hat{k}$ and radius vector is $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ then its velocity is:
(1999)
1. $-5\hat{i} + 2\hat{j} + 3\hat{k}$
2. $-5\hat{i} + 2\hat{j} - 3\hat{k}$
3. $-5\hat{i} - 2\hat{j} + 3\hat{k}$
4. $-5\hat{i} - 2\hat{j} - 3\hat{k}$
View Answer
Velocity is given by $\vec{v} = \vec{\omega} \times \vec{r}$. Evaluating the cross-product determinant yields $-5\hat{i} + 2\hat{j} + 3\hat{k}$.