Question

A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle \[{{60}^{0}}\] with vertical? (g is the acceleration due to gravity)
A) The angular acceleration of the rod will be \[\dfrac{2g}{L}\].
B) The normal reaction force from the floor on the rod will be \[\dfrac{Mg}{16}\].
C) The radial acceleration of the rod’s center of mass will be \[\dfrac{3g}{4}\].
D) The angular speed of the rod will be \[\sqrt{\dfrac{3g}{2L}}\].

Answer 1

Hint: We need to understand the different physical parameters for a long rod which undergoes a rotational motion about the axis at one of its ends to get the correct relation with them and the given information on length, mass and angle of the rod.

Complete answer:
We are given a situation where a rod of mass M and length L is placed on a floor which has sufficient friction such that the rod will not slip off even when left without holding it and landing horizontally on the floor. We can consider this lower end to be attached to the floor due to this reason.

Now, let us consider when the rod is falling on the floor. We know that the mechanical energy of the system is conserved from the journey from top to bottom. The maximum potential energy at top becomes the maximum kinetic energy when hitting the floor as –
\[\begin{align}
  & PE=KE \\
 & \Rightarrow PE=\dfrac{1}{2}{{I}_{0}}{{\omega }^{2}} \\
 & \therefore \dfrac{MgL}{2}\cos \theta =\dfrac{1}{2}{{I}_{0}}{{\omega }^{2}} \\
\end{align}\]
This gives mechanical energy conservation at a point between A and B.
Now, we have to find the moment of inertia of the rod about the fixed end which is half the length of the rod, which is given as –
\[{{I}_{0}}=\dfrac{M{{L}^{2}}}{3}\]
The relation between the energies will be thus given the angular velocity at the point when the rod is \[{{60}^{0}}\] to the vertical is –
\[\begin{align}
  & \dfrac{MgL}{2}\cos \theta =\dfrac{1}{2}{{I}_{0}}{{\omega }^{2}} \\
 & \Rightarrow \dfrac{MgL}{2}\cos {{60}^{0}}=\dfrac{1}{2}\dfrac{M{{L}^{2}}}{3}{{\omega }^{2}} \\
 & \Rightarrow \dfrac{g}{4}=\dfrac{1}{6}L{{\omega }^{2}} \\
 & \therefore \omega =\sqrt{\dfrac{3g}{2L}} \\
\end{align}\]
Now, we can find the radial acceleration of the rod at the same instant as –
\[\begin{align}
  & {{a}_{r}}=r{{\omega }^{2}} \\
 & \Rightarrow {{a}_{r}}={{\omega }^{2}}\dfrac{L}{2} \\
 & \Rightarrow {{a}_{r}}=\dfrac{3g}{2L}\dfrac{L}{2} \\
 & \therefore {{a}_{r}}=\dfrac{3g}{4} \\
\end{align}\]
Now, we can find the angular acceleration using the torque and the moment of inertia as –
\[\begin{align}
  & \tau ={{I}_{0}}\alpha \\
 & \Rightarrow \alpha =\dfrac{\tau }{{{I}_{0}}} \\
 & \Rightarrow \alpha =\dfrac{Mg\dfrac{L}{2}\sin \theta }{\dfrac{M{{L}^{2}}}{3}} \\
 & \Rightarrow \alpha =\dfrac{Mg\dfrac{L}{2}\sin {{60}^{0}}}{\dfrac{M{{L}^{2}}}{3}} \\
 & \therefore \alpha =\dfrac{3\sqrt{3}g}{4L} \\
\end{align}\]
Now, we can find the vertical acceleration which contributes to the net force in the system as –
\[\begin{align}
  & {{a}_{v}}=(\alpha \times r)+(r{{\omega }^{2}}\cos \theta ) \\
 & \Rightarrow {{a}_{v}}=(\dfrac{3\sqrt{3}g}{4L}\dfrac{L}{2}\sin {{60}^{0}})+\dfrac{L}{2}(\dfrac{3g}{2L})\cos {{60}^{0}} \\
 & \Rightarrow {{a}_{v}}=\dfrac{9}{4}g+\dfrac{3}{8}g \\
 & \therefore {{a}_{v}}=\dfrac{15}{16}g \\
\end{align}\]
Now, the vertical forces can be given as –
\[\begin{align}
  & F=M{{a}_{v}} \\
 & \Rightarrow Mg-N=M{{a}_{v}} \\
 & \Rightarrow N=M(g-{{a}_{v}}) \\
 & \Rightarrow N=M(g-\dfrac{15}{16}g) \\
 & \therefore N=\dfrac{Mg}{16} \\
\end{align}\]
So, we get the important parameters involved in this problem.

The correct answers are options B, C and D.

Note:
The rod is falling from its initial position to the floor without slipping. Therefore, we can understand that there is no net force acting in the horizontal direction and so we can find the net force using the vertical acceleration component to find the normal.