Kinetics of Particles Questions and Answers

Answer: B

The speed at which a car needs no friction at all on a banked curve is \(v = \sqrt{rg\tan\theta}\). At this speed the tyres do not rub sideways against the road, which is the optimum speed to avoid wear and tear. With r = 200 m, g = 10 m/s\(^2\) and \(\tan 10^\circ = 0.1763\):

The given coefficient of friction 0.2 is not used, because it only determines the maximum safe speed (about 27 m/s), not the optimum one. The other options do not follow from either speed.

Answer: A

The Coriolis acceleration is 1.2 \(\text{m/s}^{2}\).

For a body moving with velocity \(v\) relative to a frame rotating at \(\omega\), with \(v\) perpendicular to the rotation axis, the Coriolis component is \(2\omega v\):

\[ \begin{aligned} a_{c} &= 2\omega v = 2 \times 3 \times 0.2 \\ &= 1.2\ \text{m/s}^{2} \end{aligned} \]

0.6 omits the factor 2, 2.4 doubles it again, and 0.3 is neither. The direction is perpendicular to the slot, but the question asks only for the magnitude.

Answer: D

The tension between A and B is \(F(m_{B} + m_{C})/(m_{A} + m_{B} + m_{C})\).

The whole train accelerates together at

\[ a = \frac{F}{m_{A} + m_{B} + m_{C}} \]

The coupling between A and B is the only horizontal force on the rear part of the train, coaches B and C together, so it must give them this acceleration:

\[ \begin{aligned} T_{AB} &= (m_{B} + m_{C})a \\ &= \frac{F(m_{B} + m_{C})}{m_{A} + m_{B} + m_{C}} \end{aligned} \]

F would be the tension only if B and C were massless. \(Fm_{A}/(\ldots)\) is the net force on A, and \(Fm_{B}/(\ldots)\) is the net force on B alone.

Answer: B

The final angular speed is 1 rad/s.

The turntable is massless and its bearings are frictionless, so no external torque acts about the axis and angular momentum is conserved. Friction between the disks is internal.

\[ \begin{aligned} I\omega_{0} &= (I + I)\omega_{f} \\ \omega_{f} &= \frac{\omega_{0}}{2} = \frac{2}{2} = 1\ \text{rad/s} \end{aligned} \]

The mass and radius are not needed, since both disks are identical and their moments of inertia cancel. 2 rad/s would mean the second disk added no inertia, 4 rad/s would violate conservation, and 0.15 rad/s has no basis. Kinetic energy is not conserved: half of it is lost to friction.

Answer: B

The angular velocity is \(\sqrt{\frac{2g}{L}}\).

The hinge is frictionless and the rod massless, so mechanical energy is conserved. Swinging from horizontal to vertical, the mass falls a height \(L\), and in the vertical position it moves on a circle of radius \(L\) with speed \(\omega L\):

\[ \begin{aligned} mgL &= \tfrac{1}{2}m(\omega L)^{2} \\ \omega^{2} &= \tfrac{2g}{L} \\ \omega &= \sqrt{\tfrac{2g}{L}} \end{aligned} \]

\(\sqrt{g/L}\) drops the factor 2 from the kinetic energy, \(2\sqrt{g/L}\) is the speed for a fall of \(2L\), and \(\frac{1}{2}\sqrt{g/L}\) has no consistent derivation.