An unbanked road has a radius of curvature equal to \(30\text{ m}\). The maximum speed at which a car can make a turn if the coefficient of friction is \(0.75\), is
The maximum safe speed on an unbanked circular curve is \(v_{\max} = \sqrt{\mu g r}\). Substituting the values, \(v_{\max} = \sqrt{0.75 \times 10 \times 30} = \sqrt{225} = 15\text{ m/s}\).
A gramophone record is revolving with an angular velocity \( \omega \). A coin is placed at a distance \( r \) from the center of the record. The static coefficient of friction is \( \mu \). The coin will revolve with the record if:
(2010 Pre)
For the coin to revolve without slipping, the static friction must provide the necessary centripetal force. Required centripetal force \( F_{\text{c}} = mr\omega^2 \). Maximum static friction \( f_{\text{s,max}} = \mu mg \). So, \( mr\omega^2 \le \mu mg \). This simplifies to \( r\omega^2 \le \mu g \), or \( r \le \frac{\mu g}{\omega^2} \).
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)
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 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)
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)
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)
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)
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)
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, \(W = 0\).