Question

A 75 g, 30-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 10 g ball of clay traveling horizontally at 2.5 m/s hits and sticks to the very
bottom tip of the rod. To what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?

Short answer

The maximum angle will be 67^o

Step-by-step solution

Given information

Mass of the axle=75 g =0.075kg
Length of the axle = 30 cm =0.3 m
Mass of clay ball =10 g = 0.010 kg
Velocity of the clay ball =2.5 m/s

Explanation

First draw the diagram as below to understand and solve the problem

Use the law of conservation of momentum

Initial angular momentum about O is given as ,

$L=mv\frac{l}{2}+I{\omega }_i......................................\left(1\right)$

Initial at rest so ω_i=0

$\Rightarrow L=mv\frac{l}{2}+0....................................\left(2\right)$

Substitute the values

$\Rightarrow L=(0.01kg)(0.15m/s)(2.5m)=0.00375kg{m}^2/s................................\left(3\right)$

Momentum after ball hits

$L^{\text{'}}=\left(\frac{M{l}^2}{12}+\frac{M{l}^2}{4}\right){\omega }^{\text{'}}...........................\left(4\right)$

Substitute the values

$$\begin{gathered} L^{\text{'}}=\left(\frac{0.075kg(0.3m{)}^2}{12}+\frac{(0.01kg)(0.3m{)}^2}{4}\right){\omega }^{\text{'}} \\ \Rightarrow 0.00375kg{m}^2/s=0.00079{\omega }^{\text{'}}kg{m}^2 \\ \Rightarrow {\omega }^{\text{'}}=\frac{0.00375kg{m}^2/s}{0.00079kg{m}^2}=4.7\mathrm{rad}/\sec \dots \dots \dots \dots \dots \dots \dots \left(5\right) \end{gathered}$$

Now use the law of conservation of energy

$\frac{1}{2}I{\omega }^{\text{'}2}=(M+m)g\frac{l}{2}(1-\cos \theta ).........................\left(6\right)$

Substitute values in equation (6)

$\Rightarrow \frac{1}{2}(0.00079kg{m}^2)(4.7/s{)}^2=(0kg+0.01kg\left)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{0.30m}{2}\right)\right(1-\cos \theta )$

Now find the angle

$$\begin{gathered} \Rightarrow 0.0088=0.0147(1-\cos \theta ) \\ \Rightarrow 0.59=(1-\cos \theta ) \\ \Rightarrow \cos \theta =1-0.59 \\ \Rightarrow \theta ={\cos }^{-1}(0.4)=67° \end{gathered}$$