A car of mass \(1000\text{ kg}\) negotiates a banked curve of radius \(90\text{ m}\) on a frictionless road. If the banking angle is \(45^\circ\), the speed of the car is:
(2012 Pre)
1. \(20\text{ m/s}\)
2. \(30\text{ m/s}\)
3. \(5\text{ m/s}\)
4. \(10\text{ m/s}\)
View Answer
For an ideal frictionless banked curve, the optimum speed is given by \(v = \sqrt{gRtan\theta}\). Given R = 90 m, \(\theta = 45^\circ\), and assuming \(g = 10\text{ m/s}^2\), \(v = \sqrt{10 \times 90 \times tan 45^\circ} = \sqrt{900 \times 1} = 30\text{ m/s}\).
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 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 particle of mass \(m\) is tied to a string of length \(l\) and whirled into a horizontal plane. If tension in the string is \(T\) then the speed of the particle will be:
(1999)
1. \(\sqrt{\frac{Tl}{m}}\)
2. \(\sqrt{\frac{2Tl}{m}}\)
3. \(\sqrt{\frac{3Tl}{m}}\)
4. \(\sqrt{\frac{Tl}{2m}}\)
View Answer
Concept: Centripetal force is provided by the tension in the string.
Formula: Centripetal force \(F_c = \frac{mv^2}{l}\). Here, \(F_c = T\).
Solution: \(T = \frac{mv^2}{l}\). Rearranging for \(v\), we get \(v^2 = \frac{Tl}{m}\), so \(v = \sqrt{\frac{Tl}{m}}\).
When milk is churned, cream gets separated due to:
(1999)
1. Centripetal force
2. Centrifugal force
3. Frictional force
4. Gravitational force
View Answer
Concept: Understanding apparent forces in a non-inertial rotating frame of reference.
Solution: When milk is churned, the denser skim milk experiences a larger centripetal force and moves towards the outer edge, while the less dense cream moves towards the center of rotation due to centrifugal force. Thus, they separate.
When a body moves with a constant speed along a circle:
(1994)
1. No work is done on it
2. No acceleration is produced in it
3. Its velocity remains constant
4. No force acts on it
View Answer
In uniform circular motion, the centripetal force is always perpendicular to the instantaneous displacement. Work \(W = F.d cos\theta\). Since \(\theta = 90^{\circ}\), \(cos\theta = 0\). Therefore, \(W = 0\).