Question

A meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor is assuming that the end at the floor does not slip is (\[g = 9.8\,{\text{m/}}{{\text{s}}^2}\])
A. \[3.2\,{\text{m/s}}\]
B. \[5.4\,{\text{m/s}}\]
C. \[7.6\,{\text{m/s}}\]
D. \[9.2\,{\text{m/s}}\]

Answer 1

Hint: Use the formula for the potential energy and rotational kinetic energy of an object. Also use the formula for moment of inertia of the rod about an axis passing through one end of the rod. Use the law of conservation of energy for the rod at the vertical position and when it falls on the ground. Using the relation between the linear and angular velocity, calculate the linear velocity of the other end of the rod.

Formulae used:
The potential energy \[U\] of an object is
\[U = mgh\] …… (1)
Here, \[m\] is the mass of the object, \[g\] is acceleration due to gravity and \[h\] is the height of the object from the ground.
The rotational kinetic energy \[K\] of an object is
\[K = \dfrac{1}{2}I{\omega ^2}\] …… (2)
Here, \[I\] is the moment of inertia of the object and \[\omega \] is angular velocity of the object.
The moment of inertia \[I\] of the rod about an axis passing through its one end is
\[I = \dfrac{{M{L^2}}}{3}\] …… (3)
Here, \[M\] is the mass of the rod and \[L\] is the length of the rod.
The linear velocity \[v\] of an object is given by
\[v = R\omega \] …… (4)
Here, \[R\] is the radius of the circular path and \[\omega \] is the angular velocity of the object.

Complete step by step answer:
We have given that a meter stick is held vertically with one end of the stick on the floor and the other end is allowed to fall on the ground. We have asked to calculate the speed of the other end. We know that the length of the meter stick is \[1\,{\text{m}}\].
\[L = 1\,{\text{m}}\]
The height of the centre of mass of the stick from the ground is at a distance half of the length of the stick.
\[h = \dfrac{L}{2}\]
The potential energy of the centre of mass of the stick is given by
\[U = mg\left( {\dfrac{L}{2}} \right)\]
\[ \Rightarrow U = \dfrac{{mgL}}{2}\]
The rotational kinetic energy of the centre of mass of the stick is given by
\[K = \dfrac{1}{2}I{\omega ^2}\]

According to the law of conservation of energy, the potential energy of centre of mass of the meter stick is equal to the rotational kinetic energy of the meter stick.
\[U = K\]
\[ \Rightarrow \dfrac{{mgL}}{2} = \dfrac{1}{2}I{\omega ^2}\]
\[ \Rightarrow mgL = I{\omega ^2}\]
Substitute \[\dfrac{{m{L^2}}}{3}\] for \[I\] in the above equation.
\[ \Rightarrow mgL = \dfrac{{m{L^2}}}{3}{\omega ^2}\]
\[ \Rightarrow g = \dfrac{L}{3}{\omega ^2}\]
\[ \Rightarrow \omega = \sqrt {\dfrac{{3g}}{L}} \]
The linear velocity of the other end of the meter stick is given by
\[v = L\omega \]

Substitute \[\sqrt {\dfrac{{3g}}{L}} \] for \[v = L\omega \] in the above equation.
\[v = L\sqrt {\dfrac{{3g}}{L}} \]
\[ \Rightarrow v = \sqrt {3gL} \]
Substitute \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] and \[1\,{\text{m}}\] for \[L\] in the above equation.
\[ \Rightarrow v = \sqrt {3\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {1\,{\text{m}}} \right)} \]
\[ \Rightarrow v = \sqrt {29.4} \]
\[ \therefore v = 5.4\,{\text{m/s}}\]
Therefore, the velocity of the other end is \[5.4\,{\text{m/s}}\].

Hence, the correct option is B.

Note: The students should not forget to calculate the potential energy of centre of mass of the meter stick which is at half length of the meter stick because the meter stick in its vertical position is in equilibrium as the centre of mass of the meter stick is in equilibrium. If one calculates the potential energy of the meter stick at the upper end of the meter stick then it will give an incorrect answer.