The figure shows a pendulum of length $\mathrm{L}=1.25\mathrm{m}$. Its bob (which effectively has all the mass) has speed $v_0$when the cord makes an angle ${\mathrm{\theta }}_0=40.0°$with the vertical.

1. What is the speed of the bob when it is in its lowest position if ${\mathrm{V}}_0=8.00\text{m/s}$? What is the least value that $v_0$can have if the pendulum is to swing down and then up
2. To a horizontal position.
3. To a vertical position with the cord remaining straight?
4. Do the answers to (b) and (c) increase, decrease, or remain the same if they areincreased by a few degrees?

1. Length of string,$L=125m$
2. Angle of pendulum with vertical,$\theta ={40}^o$

Using the energy conservation law, find the velocity at the lowest point. So, according to the condition that least velocity is given, the string should be straight. Find the velocity at the horizontal and vertical positions. According to the law of energy conservation, energy can neither be created nor destroyed.

Formulae:

$$\begin{gathered} \text{i)Energy = PE + KE = mgh +}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}} \\ \text{ii) W = mg}\backslash \mathrm{hfill} \\ \text{iii) Centripetal force, F =}\frac{{\text{mv}}^{\text{2}}}{\text{r}} \end{gathered}$$

where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, x is displacement, r is radius, and W is work done.

From the geometry of the above figure,

$h=L-L\left(\mathrm{cm}\theta \right)$

$h=L\left(1-\left(\cos \theta \right)\right)$

Also,

$$\begin{gathered} T=mg \\ =W \end{gathered}$$

At the lowest position, the object has only kinetic energy. So, by the law of conservation of energy,

$$\begin{gathered} {\left(KE+PE\right)}_{any}={\left(KE+PE\right)}_{lowest} \\ \frac{1}{2}m{v}_0^2+mgh=\frac{1}{2}m{v}^2+0 \\ \frac{1}{2}m{v}_0^2+mgL\left(1-\left(\cos \theta \right)\right)=\frac{1}{2}m{v}^2 \\ {v}_0^2+2gL\left(1-\left(\cos \theta \right)\right)=v^2 \\ v=\sqrt{v_0^2+2gL\left(1-\left(\cos \theta \right)\right)} \\ \text{After plugging the values,}\backslash \mathrm{hfill} \\ v=\sqrt{{\left(8\right)}^2+2\left(9.8\right)\left(1.25\right)\left(1-\cos 40\right)} \\ v=8.38\text{m/s} \end{gathered}$$

Hence, speed of the bob when it is at the lowest position and has speed is $8.38\text{m/s}$.

Now, find the speed of the cord at the highest point. And, there is no kinetic energy at this point. So,

$$\begin{gathered} {\left(KE+PE\right)}_{any}={\left(KE+PE\right)}_{Horizontal} \\ {\left(KE+PE\right)}_{any}={\left(0+PE\right)}_{horizontal} \\ \frac{1}{2}m{v}_0^2+mgh=mgL \\ \frac{1}{2}m{v}_0^2+mgL\left(1-\cos \theta \right)=mgL \\ {v}_0^2=2gL-2gL\left(1-\cos \theta \right) \\ {v}_0=\sqrt{2gL-2gL+2gL\cos \theta )} \\ {v}_0=\sqrt{2gL\cos \theta )} \\ {v}_0=\sqrt{2\left(9.8\right)\left(1.25\right)\left(\cos 40\right)} \\ {v}_0=4.33\text{m/s} \end{gathered}$$

Hence, the least value that $v_0$can have if the pendulum is swung down and then up to the horizontal position is role="math" localid="1663124831697" $4.33\text{m/s}$.

Apply Newton's third law for the least velocity with which thestring should be straight.

From the figure,

$T=mg$

And at the highest point, role="math" localid="1663124974743" $\theta =180°$

Tension is force, and force is equal to centripetal force.

$$\begin{gathered} F=\frac{m{v}_t^2}{r} \\ T=\frac{m{v}_t^2}{r}=mg \\ m{v}_t^2=mgr \\ =mgL \end{gathered}$$

Again, recall the law conservation of energy,

$$\begin{gathered} {\left(KE+PE\right)}_{any}={\left(KE+PE\right)}_{top} \\ \frac{1}{2}m{v}_0^2+mgL\left(1-\cos \theta \right)=\frac{1}{2}m{v}_t^2+mg\left(1-\cos 180\right) \\ {v}_0^2+2gL\left(1-\cos 40\right)=\left(gL\right)+2g\left(1-\cos 180\right) \\ {v}_0^2+2gL-2gL\cos 40=gL+2g-2g\cos 180 \\ {v}_0^2=gL+2g\left(1-\cos 180\right)+2gL\left(\cos 40-1\right) \end{gathered}$$

After plugging all the values properly,

role="math" localid="1663125143709" $v_0^2=55.51$

$v_0=7.4\text{5m/s}$

Hence, the least value that $v_0$can have if the pendulum is swung down and then upto the vertical position is$7.4\text{5m/s}$.

From the derived equation of,

For part (b),$v_0=\sqrt{\left(2gL\cos \theta \right)}$

&

For part (c),$v_0^2=gL+2g\left(1-\cos 180\right)+2gL\left(\cos {\theta }_0-1\right)$

The values of cosine are going to decrease as we increase the angle.

So, the answers in parts (b) and (c) will decrease if the value of ${\theta }_0$is increased.

Hence, the effect on answers of (b) and (c) if the angle is increased is that the answers will decrease.