A massless rigid rod of length Lhas a ball of massm attached to one end. The other end is pivoted in such a way that the ball will move in a vertical circle. First, assume that there is no friction at the pivot. The system is launched downward from the horizontal position A with initial speed ${\mathit{v}}_{\mathbf{0}}$. The ball just barely reaches point D and then stops. (a) Derive an expression for ${\mathit{v}}_{\mathbf{0}}$in terms of L, m, and g. (b) What is the tension in the rod when the ball passes through B? (c) A little grit is placed on the pivot to increase the friction there. Then the ball just barely reaches C when launched from A with the same speed as before. (d) What is the decrease in the mechanical energy by the time the ball finally comes to rest at B after several oscillations?

A massless rigid rod of length L has a ball of mass m attached to one end and is launched at an initial speed $v_0$and then stops having vertical motion.

By calculating the mechanical energy at each point and conservation of energy, we get the equation for the velocity of the ball. Using conservation of energy and energy at point B, we can find the tension in the rod. Also, using the conservation of energy we can find the answers for parts (c) and (d).

Formulae:

The force due to Newton’s second law, F = ma (1)

The kinetic energy of the body in motion, $\mathrm{KE}=\frac{1}{2}m{v}^2$ (2)

The potential energy of a body at a height, PE = mgh (3)

The centripetal acceleration of the body, $a=\frac{v^2}{r}$ (4)

Mechanical Energy at point A using equation (2) is given as:

${\mathrm{E}}_A=\frac{1}{2}{m}_A^2$

Mechanical Energy at point B using equations (2) and (3) is given as:

$E_B=\frac{1}{2}m{v}_B^2-mgL$

Mechanical Energy at point D using equation (3) is given as:

$E_0=mgL$

According to the law of conservation of energy, we can say that

$$\begin{gathered} E_B=E_0 \\ \frac{1}{2}m{v}_0^2=mgL \\ {v}_0=\sqrt{2gL} \end{gathered}$$

Hence, the expression for the initial speed is $\sqrt{2gL}$.

Using the law of conservation of energy, and equations (2) and (3), we can say that

$$\begin{gathered} E_A=E_B \\ \frac{1}{2}m{v}_B^2-mgL=mgL \\ {v}_B^2=4gL \\ {v}_B=\sqrt{4gL} \end{gathered}$$

The Centripetal acceleration will act upwards, when it passes point B, hence, using equation (4) in equation (1), the net force acting can be given as:

$$\begin{gathered} F_{net}=m{a}_B \\ T-mg=ma \\ T=m(a+g) \\ T=m\left(\frac{v_B^2}{r}+g\right) \\ T=m\left(\frac{4gL}{L}+g\right) \\ T=5mg \end{gathered}$$

Hence, the value of the tension is 5mg

The ball barely reaches point C, i.e. $v_C=0$,

Using the conservation of energy, we can say that the loss in Mechanical Energy of the motion is given as:

$\begin{array}{l}∆\cup =E_C-E_0\\ =0-mgL\\ =-mgL\end{array}$

Hence, the value of the decrease is -mgL

As, the ball stops at B, i.e. $v_B=0$,

Using conservation of energy, we can say that the loss in Mechanical Energy is given as:

$$\begin{gathered} ∆U=E_B-E_0 \\ =\left(\frac{1}{2}{v}_B^2-mgL\right)-\frac{1}{2}m{v}_0^2 \\ =\left(0-mgL\right)-\frac{1}{2}m\left(2gL\right) \\ =-mgL-mgL \\ =-2mgL \end{gathered}$$

Hence, the value of the decrease in the energy is -2mgL.