Chapter 8, Problem 81GP

To determine

To Find: The torque required to achieve the feat.

Answer to Problem 81GP

  $2.83\text{ N}\cdot \text{m}$

Explanation of Solution

Given data:

Mass of rock, $\mathrm{m}=0.\text{5 kg}$

Sling length, $\mathrm{l}=\text{1}.\text{5 m}$

Speed, $\mathrm{N}=\text{ 12}0\text{ rpm}$

Time interval, $\mathrm{t}=\text{5}\text{s}$

Formula used:

Write the expression for the torque.

  $\mathrm{\tau }=\mathrm{I}\mathrm{\alpha }$

Here, $\mathrm{I}$ is the moment of inertia,

  $\mathrm{\alpha }$ is the angular acceleration.

Angular acceleration can be written as,

  $\mathrm{\alpha }=\left(\frac{{\mathrm{\omega }}_{\mathrm{f}}-{\mathrm{\omega }}_{\mathrm{i}}}{\mathrm{t}}\right)$

  ${\mathrm{\omega }}_{\mathrm{i}}$ is the initial angular velocity,

  ${\mathrm{\omega }}_{\mathrm{f}}$ is the final angular velocity.

Calculation:

The torque required is given by the formula,

  $\mathrm{\tau }=\mathrm{I}\mathrm{\alpha }$

  $\mathrm{\tau }=\mathrm{I}\left(\frac{{\mathrm{\omega }}_{\mathrm{f}}-{\mathrm{\omega }}_{\mathrm{i}}}{\mathrm{t}}\right)$

  ${\mathrm{\omega }}_{\mathrm{i}}=0$

  $\mathrm{\tau }=\mathrm{m}.{\mathrm{r}}^2.\frac{2\mathrm{\pi }\mathrm{N}}{\mathrm{t}}$

  $\mathrm{\tau }=0.5\times {1.5}^2\times \frac{2\mathrm{\pi }\times 120}{5\times 60}$

  $\mathrm{\tau }=2.83\text{ N}\cdot \text{m}$

Conclusion:

The torque required to achieve David’s feat is $2.83\text{ N}\cdot \text{m}$ .