Question

A pendulum with a cord of length $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{m}}$swings in a vertical plane (Fig. P4.70).When the pendulum is in the two horizontal positions $\mathit{\theta }\mathbf{=}{\mathbf{90}}^{\mathbf{\circ }}$and localid="1663633590926" $\mathit{\theta }\mathbf{=}{\mathbf{270}}^{\mathbf{\circ }}$, its speed is $\mathbf{5}\mathbf{}\mathbf{\text{m/s}}$. Find the magnitude of (a) the radial acceleration and (b) the tangential acceleration for these positions. (c) Draw vector diagrams to determine the direction of the total acceleration for these two positions. (d) Calculate the magnitude and direction of the total acceleration at these two positions.

Short answer

1. The magnitude of radial acceleration is $25.0\mathrm{m}/{\mathrm{s}}^2$.
2. The tangential acceleration is$9.81\mathrm{m}/{\mathrm{s}}^2$.
3. The vector diagrams are shown below in the solution.
4. The magnitude and direction of the total acceleration at these two positions $26.9\mathrm{m}/{\mathrm{s}}^2$and $21.4^{\circ }$below horizontal.

Step-by-step solution

Radial acceleration.

In a uniform circular when the acceleration of the object is along the radius, directed towards the centre it is called radial acceleration.

${\mathit{a}}_{\mathbf{r}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here role="math" localid="1663633117278" $\mathit{v}$is the velocity,$\mathit{r}$ is the radius, ${\mathit{a}}_{\mathbf{r}}$is the radial acceleration.

Find the radial acceleration.

(a)

Given $v=5.00\mathrm{m}/\mathrm{s}$and $r=1\text{m}$.

Hence, the magnitude of radial acceleration is given by

role="math" localid="1663633241541" $\begin{array}{c}{a}_r=\frac{v^2}{r}\\ =\frac{{(5.00\mathrm{m}/\mathrm{s})}^2}{(1.00\mathrm{m})}\\ =25.0\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the magnitude of radial acceleration is $25.0\mathrm{m}/{\mathrm{s}}^2$.

Find the tangential acceleration.

(b)

Since, at this point the pendulum is horizontal, the tangential acceleration will come from the vertical force. Now at this point only vertical force acting on the ball is gravity. Hence, the tangential acceleration is the acceleration due to gravity, which is

$\begin{array}{c}{a}_t=g\\ =9.81 \mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the tangential acceleration is $9.81 \mathrm{m}/{\mathrm{s}}^2$.

Vector diagrams to determine the direction of the total acceleration when θ=90∘.

(c)

 Step 5: The magnitude of the total acceleration.

(d)

The magnitude of the total acceleration is given by

$\begin{array}{c}a=\sqrt{a_r^2+a_t^2}\\ =\sqrt{{\left(25.0\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}^2}\\ =26.9\mathrm{m}/{\mathrm{s}}^2\end{array}$

And the direction of the total acceleration is given by

$\begin{array}{c}\varphi =\frac{a_t}{a_r}\\ =\arctan \left\{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(25.0\mathrm{m}/{\mathrm{s}}^2\right)}\right\}\\ =21.4°\end{array}$

Hence, the magnitude of the total acceleration is$26.9\mathrm{m}/{\mathrm{s}}^2$and the direction is$21.4°$below the horizontal.