Question

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length $\mathbf{55}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}$. At the instant it makes an angle of $\mathbf{\text{35.0°}}$with the vertical as it falls, what are

(a) the radial acceleration of the top, and

(b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.)

(c) At what angle$\mathbf{\text{θ}}$ is the tangential acceleration equal to g?

Short answer

1. Radial acceleration of the top is$5.32\text{rad}/{\text{s}}^2$
2. Tangential acceleration of the top is$8.43\text{\hspace{0.17em}m}/{\text{s}}^2$
3. The angle $\mathrm{\theta }$at which tangential acceleration is equal to g is$41.8°$.

Step-by-step solution

Given

Length of the Chimney is$,\text{L}=55.0\text{\hspace{0.17em}m}$

$\mathrm{\theta }=35.0°$

Understanding the concept

Using the law of conservation of energy, we can findtheangular speed of the top. From this, we can easily find its radial acceleration usingthecorresponding relation. Fromtheangular speed, we can find its angular acceleration. Using this, we can find its tangential acceleration usingthecorresponding formula. Then insertingthegiven condition in the expression for the tangential acceleration, we can find the angle at that condition.

Formulae:

The M.I of the rod about an axis passing through it’s one end is,$\text{I}=\frac{1}{3}{\text{ml}}^2$

${\text{E}}_{\mathrm{i}}={\text{E}}_{\mathrm{f}}$

${\text{a}}_r=\text{r}{\omega }^2$

${\text{a}}_t=\text{r}\alpha$

(a) Calculate the radial acceleration of the top

Let m be the mass of the chimney.

Initially, Its center of mass is at L/2. After falling, let’s assume that it is at l.

Since the chimney makes an angle$\mathrm{\theta }$with verticle,

$\text{l}=\frac{\text{L}}{\text{2}}\text{cos}\mathrm{\theta }$

The M.I of the chimney is,

$\text{I}=\frac{1}{3}{\text{mL}}^2$

Chimney is released from rest. So it’s initial K.E is zero.

According to the conservation of energy,

${\text{E}}_{\mathrm{i}}={\text{E}}_{\mathrm{f}}$

$\Rightarrow {\text{K.E}}_{\mathrm{i}}+{\text{P.E}}_{\mathrm{i}}={\text{K.E}}_{\mathrm{f}}+{\text{P.E}}_{\mathrm{f}}$

In this case,

$0+\frac{\text{mgL}}{2}=\text{mgl}+\frac{1}{2}\text{I}{\mathrm{\omega }}^2+0$

$\Rightarrow \frac{\text{mgL}}{2}=\frac{\text{mgL}}{2}\text{cos}\theta +\frac{1}{2}\left(\frac{1}{3}{\text{mL}}^2\right){\omega }^2$

$\therefore {\omega }^2=\frac{6}{{\text{mL}}^2}\left(\frac{\text{mgL}}{2}-\frac{\text{mgL}}{2}\text{cos}\theta \right)$

$\Rightarrow {\mathrm{\omega }}^2=\frac{6}{\text{L}}\left(\frac{\text{g}}{2}-\frac{\text{g}}{2}\text{cos}\mathrm{\theta }\right)$

$\Rightarrow \omega =\sqrt{\frac{6}{\text{L}}\left(\frac{\text{g}}{2}-\frac{\text{g}}{2}\text{cos}\theta \right)}$

$\Rightarrow \mathrm{\omega }=\sqrt{\frac{6}{55}\left(\frac{9.8}{2}-\frac{9.8}{2}\text{cos}(35.0°)\right)}$

$\Rightarrow \mathrm{\omega }=0.3109~0.311\text{\hspace{0.17em}rad}/\text{s}$

Radial acceleration of the top is

${\text{a}}_{\mathrm{r}}=\text{L}{\mathrm{\omega }}^2=55{\left(0.3109\right)}^2$

$\Rightarrow {\text{a}}_{\mathrm{r}}=5.316~5.32\text{rad}/{\text{s}}^2$

Therefore, radial acceleration of the top is$5.32\mathrm{rad}/{\mathrm{s}}^2.$

(b) Calculate the tangential acceleration of the top

We have

${\omega }^2=\frac{6}{\text{L}}\left(\frac{\text{g}}{2}-\frac{\text{g}}{2}\text{cos}\theta \right)$

$\Rightarrow {\omega }^2=\frac{3\text{g}}{\text{L}}(1-\text{cos}\theta )$

Differentiating with respect to t, we get

$2\omega \frac{\text{d}\omega }{\text{dt}}=\frac{3\text{g}}{\text{L}}\frac{\text{d}\left(\text{cos}\theta \right)}{\text{dt}}$

But,

$\frac{\text{d}\mathrm{\omega }}{\text{dt}}=\mathrm{\alpha }\text{and\hspace{0.17em}}\frac{\text{d}\mathrm{\theta }}{\text{dt}}=\mathrm{\omega }$

$\therefore 2\mathrm{\omega \alpha }=\frac{3\text{g}}{\text{L}}\mathrm{\omega }\text{sin}\mathrm{\theta }$

$\Rightarrow \mathrm{\alpha }=\frac{3\text{g}}{2\text{L}}\text{sin}\mathrm{\theta }$

Tangential acceleration of the top is

${\text{a}}_{\mathrm{t}}=\text{L}\mathrm{\alpha }$

$\Rightarrow {\text{a}}_{\mathrm{t}}=\frac{3\text{g}}{2}\text{sin}\mathrm{\theta }$

$\Rightarrow {\text{a}}_{\mathrm{t}}=\frac{3\left(9.8\right)}{2}\text{sin}(35.0°)$

$\Rightarrow a_t=8.43\text{m}/{\text{s}}^2$

Therefore, tangential acceleration of the top is $8.43\mathrm{m}/{\mathrm{s}}^2.$

(c) Calculate the angle  θat which the tangential acceleration equal to g

If${\text{a}}_{\mathrm{t}}=\text{g},\text{then}$

$\frac{3\text{g}}{2}\text{sin}\mathrm{\theta }=\text{g}$

$\Rightarrow \frac{3}{2}\text{sin}\theta =1$

$\Rightarrow \text{sin}\mathrm{\theta }=\frac{2}{3}$

$\Rightarrow \theta =41.8°$

Therefore, the angle $\theta$at which tangential acceleration is equal to g is$41.8°$