Question

80. A common demonstration, illustrated in Figure P10.80, consists of a ball resting at one end of a uniform board of length ,$\mathit{I}$
that is hinged at the other end and elevated at an angle$\mathit{\theta }$. A light cup is attached to the board at${\mathit{r}}_{\mathbf{c}}$.so that it will catch the ball when the support stick is removed suddenly. (a) Show that the ball will lag behind the falling board when$\mathit{\theta }$is less than.$\mathbf{35}\mathbf{.}{\mathbf{3}}^{\mathbf{\circ }}$

(b) Assuming the board is$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{m}$long and is supported at this limiting angle, show that the cup must be$\mathbf{18}\mathbf{.}\mathbf{4}\mathit{c}\mathit{m}$from the moving end.

Short answer

The solution is

a)$\theta \le 35.3^{\circ }$

b) the cup should be$(1.00m-0.816m)=0.184m$ from the moving end

Step-by-step solution

Given Information

It is given that

Length of the board =$I$$l$

Light cup is attached atlocalid="1663656103738" $r_c$

Take a look at the free body diagram below

Balancing the torque

For a board that is just getting started moving,

$\begin{array}{l}\mathbf{\Sigma }\mathbf{\tau }\mathbf{=}\mathbf{I}\mathbf{\alpha }\\ \mathbf{m}\mathbf{g}\mathbf{\left(}\frac{\mathbf{l}}{\mathbf{2}}\mathbf{\right)}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{m}{\mathbf{l}}^{\mathbf{2}}\mathbf{\right)}\mathbf{\alpha }\end{array}$

There will be an angular acceleration

$\alpha =\frac{3}{2}\left(\frac{g}{l}\right)\cos \theta$

The end's tangential acceleration will be

$\begin{array}{c}{a}_f=l\alpha \\ =\frac{3}{2}g\cos \theta \end{array}$

The vertical component is referred to as

$\begin{array}{c}{a}_y=a_{1}cos\theta \\ =\frac{3}{2}gco{s}^2\theta \end{array}$

The board will pull ahead of the falling ball if this quantity is bigger than g.

Check whether the ball is lagging

a)

$\frac{3}{2}g{\cos }^{2}g\ge g$

This gives us

$co{s}^2\theta \ge \frac{2}{3}$

So we get

$cos\theta \ge \sqrt{\left(\frac{2}{3}\right)}$

Or

$\theta \le 35.3^{°}$

Distance between the cup and hinged end

b) In this section, we are told that

$\begin{array}{l}l=1.00m\\ \theta =35.3^{°}\end{array}$

As a result, if, the cup will land beneath the local's released point$r_c=lcos\theta$

$\begin{array}{c}{r}_c=(1.00\text{m})\sqrt{\left(\frac{2}{3}\right)}\\ =0.816\text{m}\end{array}$

As a result, the cup should be $(1.00m-0.816m)=0.184m$from the moving end