Question

A spherical asteroid with radius\(r = 123\;{\rm{m}}\)and mass\(M = 2.25 \times {10^{10}}\;{\rm{kg}}\)rotates about an axis at four revolutions per day. A “tug” spaceship attaches itself to the asteroid’s south pole (as defined by the axis of rotation) and fires its engine, applying a force F tangentially to the asteroid’s surface as shown in Fig. 8–65. If\(F = 285\;{\rm{N}}\)how long will it take the tug to rotate the asteroid’s axis of rotation through an angle of 5.0° by this method?

Short answer

The time taken by the tug to rotate the asteroid’s axis of rotation is 27 h.

Step-by-step solution

Identification of the given data

The radius of the asteroid is\(r = 123\;{\rm{m}}\).

The mass of the asteroid is\(m = 2.25 \times {10^{10}}\;{\rm{kg}}\).

The magnitude of force applied to the asteroid is\(F = 285\;{\rm{N}}\).

The angle of rotation is \(\theta = \frac{5}{{360}} \times 2\pi \;{\rm{rad}} = \frac{\pi }{{36}}\;{\rm{rad}}\).

The angular speed of the asteroid is \(\omega = \frac{{8\pi \;{\rm{rad}}}}{{86400\;{\rm{s}}}} = \frac{{8\pi }}{{86400}}\;{\rm{rad/s}}\).

Definition of angular momentum

Angular momentum is the rotational equivalent of the linear momentum of a body. It is expressed as the product of the moment of inertia and angular velocity.

\(L = I\omega \) … (i)

Newton’s second law for rotational dynamics

The rotational analog of Newton’s second law of motion states that the net torque acting on a body is equal to the time rate of change of angular momentum. It is given as:

\(\tau = \frac{{\Delta L}}{{\Delta t}}\)

Calculation of the time taken by the tug to rotate the asteroid’s axis of rotation

The spaceship applies a torque on the asteroid which changes the direction of its angular momentum. Also, torque can be written as the time rate of change of angular momentum.

\(\begin{aligned}{c}\tau = \frac{{\Delta L}}{{\Delta t}}\\ = \frac{{L\Delta \theta }}{{\Delta t}}\\\Delta t = \frac{{L\Delta \theta }}{\tau }\end{aligned}\) … (ii)

The moment of inertia of the asteroid (solid sphere) is\(I = \frac{2}{5}m{r^2}\).

Substitute equation (i) in (ii).

\(\begin{aligned}{c}\Delta t = \frac{{I\omega \Delta \theta }}{\tau }\\ = \frac{{I\omega \Delta \theta }}{{rF}}\\ = \frac{{\left( {\frac{2}{5}m{r^2}} \right)\omega \Delta \theta }}{{rF}}\\ = \frac{{\frac{2}{5}mr\omega \Delta \theta }}{F}\end{aligned}\) … (iii)

Substituting the known numerical values in equation (iii), you get:

\(\begin{aligned}{c}\Delta t = \frac{{\frac{2}{5}\left( {2.25 \times {{10}^{10}}\;{\rm{kg}}} \right)\left( {123\;{\rm{m}}} \right)\left( {\frac{{8\pi }}{{86400}}\;{\rm{rad/s}}} \right)\left( {\frac{\pi }{{36}}\;{\rm{rad}}} \right)}}{{285\;{\rm{N}}}}\\ = 9.860 \times {10^4}\;{\rm{s}}\\ = 9.860 \times {10^4}\;{\rm{s}} \times \frac{{1\;{\rm{hr}}}}{{3600\;{\rm{s}}}}\\ \approx 27\;{\rm{h}}\end{aligned}\)