Question

Question:(III) A 1.80-m-long pole is balanced vertically with its tip on the ground. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? (Hint: Use conservation of energy.)

Short answer

The speed of the upper end of the pole just before it hits the ground is 7.28 m/s.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The length of the pole is\(l = 1.80{\rm{ m}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Understanding the motion of the long pole

Initially, the long pole is balanced in the vertical direction. So, it has only gravitational potential energy. It then starts falling with some velocity and hits the ground. The pole has only rotational kinetic energy on the ground.

The energy is converted from the gravitational potential energy of the pole to the rotational kinetic energy of the pole. With the conservation of energy, the speed of the upper end of the given pole can be evaluated.

The representation of the long pole

The diagram of the pole can be shown as:

Here, mg is the weight of the pole,\(\phi \)is the angle of inclination of the pole with the ground,\({F_{\rm{N}}}\)is the normal force on the pole due to ground,\({F_{{\rm{fr}}}}\)is the frictional force, v is the speed of the upper end of the pole, and I is the length of the pole.

Determination of the speed of the upper end of the pole just before it hits the ground

From the conservation of energy, the energies can be expressed as:

\(G.P.E = K.E\)

\(mg\frac{l}{2} = \frac{1}{2}I{\omega ^2}\) … (i)

Here, G.P.E is the gravitational potential energy, K.Eis the rotational kinetic energy, I is the moment of inertia, and\(\omega \)is the angular speed of the long pole. The center of mass of the pole is in the middle of the pole. So, the length in the gravitational potential energy becomes half the total length of the pole.

The moment of inertia of the long pole (rod shaped) is expressed as:

\(I = \frac{{m{l^2}}}{3}\)

The angular velocity of the pole can be expressed as:

\(\omega = \frac{v}{l}\)

Substitute the values in equation (i).

\(\begin{aligned}{c}mg\frac{l}{2} &= \frac{1}{2} \times \frac{{m{l^2}}}{3} \times {\left( {\frac{v}{l}} \right)^2}\\gl &= \frac{1}{3} \times {v^2}\\{v^2} &= 3gl\\v &= \sqrt {3gl} \end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}v &= \sqrt {3 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 1.80{\rm{ m}}} \\ &= \sqrt {52.974} {\rm{ m/s}}\\ &= 7.28{\rm{ m/s}}\end{aligned}\)

Thus, the speed of the upper end of the pole just before it hits the ground is 7.28 m/s.