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Chapter 10: Problem 64

If the earth shrinks such that its mass does not change but radius decreases to one-quarter of its original value, then one complete day will take (a) \(96 \mathrm{~h}\) (b) \(48 \mathrm{~h}\) (c) \(6 \mathrm{~h}\) (d) \(1.5 \mathrm{~h}\)

Short Answer

Expert verified
The new duration of one complete day is 1.5 hours (option d).

Step by step solution

01

Understand the problem

We need to find how the decrease in the Earth's radius affects the duration of one complete day, given that the mass of the Earth remains constant.
02

Use the formula for moment of inertia of a sphere

The moment of inertia for a solid sphere is given by the formula \( I = \frac{2}{5}MR^2 \), where \( M \) is the mass of the Earth and \( R \) is the radius. Since mass remains constant, only \( R \) will change.
03

Express the relation between radius and angular velocity

The angular momentum \( L = I \omega \), where \( \omega \) is the angular velocity, remains constant. Thus, if \( R \) is decreased to one-quarter of its original value, the new moment of inertia is \( I' = \frac{2}{5}M(\frac{R}{4})^2 = \frac{1}{16}I \).
04

Relate old and new angular velocities

Since angular momentum \( L \) is constant, \( I \omega = I' \omega' \) leads to \( I \omega = \frac{1}{16}I \omega' \). Solving for \( \omega' \), we get \( \omega' = 16\omega \). This indicates that the angular velocity increases by a factor of 16.
05

Calculate the new duration of one complete day

The original time for one complete rotation is 24 hours. Since the angular velocity has increased 16 times, the new time taken for one complete day is \( \frac{24}{16} \) hours = 1.5 hours.
06

Choose the correct answer

From the options provided, the time taken for one complete day when the Earth's radius decreases to one-quarter without a change in mass is (d) 1.5 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in understanding how objects rotate. It essentially measures how much resistance an object has to changes in its rotational speed. For a solid sphere, such as the Earth, the moment of inertia is calculated using the formula:\[I = \frac{2}{5}MR^2\]where:
  • \(I\) is the moment of inertia,
  • \(M\) represents the mass of the object,
  • and \(R\) is the radius.
This formula shows that the moment of inertia depends directly on both the mass and the square of the radius.
When the radius changes, the moment of inertia changes significantly. For example, if the radius is reduced to one-quarter of its original value, the new moment of inertia becomes \[ I' = \frac{2}{5}M\left(\frac{R}{4}\right)^2 = \frac{1}{16}I.\]
This decrease illustrates a substantial reduction in resistance to rotational change, impacting how quickly the object can spin.
Angular Velocity
Angular velocity describes how fast an object rotates or spins around an axis. An important relation in rotational dynamics is the conservation of angular momentum, which states that the total angular momentum remains constant if no external torques act on the system.
The angular momentum \(L\) is given by the product of the moment of inertia \(I\) and the angular velocity \(\omega\):\[ L = I \omega \]When the Earth shrinks, while its mass remains the same, the moment of inertia decreases, thus affecting angular velocity. By conserving angular momentum (\(I \omega = I' \omega'\)), a change in the moment of inertia directly affects angular velocity.
  • Initially: \(L = I \omega\)
  • After the change: \(L = I' \omega'\)
    Given \(I' = \frac{1}{16}I\), the angular velocity must increase to balance the equation, specifically\[\omega' = 16\omega\].
So, if the radius decreases, the object spins faster due to increased angular velocity, leading to a shorter duration for one complete rotation.
Rotational Dynamics
Rotational dynamics is the study of how forces affect the motion of rotating bodies. This area of physics underscores how angular properties, such as moment of inertia and angular velocity, interplay when objects like the Earth undergo changes.
In the case where the Earth's radius reduces to a quarter of its original size without a mass change, we observe rotational dynamics in action. The moment of inertia decreases, significantly altering the object鈥檚 rotational properties.
  • With a radius reduction: The new moment of inertia becomes smaller, indicating less resistance to change in rotation.
  • Angular Velocity Boost: Since angular momentum is conserved, \(\omega\) must increase as \(I\) decreases. This boosts the speed of rotation.
  • Time for One Rotation: The outcome is a faster spin, resulting in a new day length computed using \(\frac{24 \text{ hours}}{16} \), equating to 1.5 hours.

Overall, rotational dynamics helps explain how shifts in the radius and moment of inertia lead to changes in rotational speed and hence the period of rotation. These concepts illustrate the profound interconnectedness of physical properties in governing celestial motions.

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Most popular questions from this chapter

A spherical hollow is made in a lead sphere of radius \(R\) such that its surface touches the outside surface of the lead sphere and passes through the centre. The mass of the lead sphere before hollowing was \(M\). The force of attraction that this sphere would exert on a particle of mass \(m\) which lies at a distance \(d(>R)\) from the centre of the lead sphere on the straight line joining the centres of the sphere and the hollow is (a) \(\frac{G M m}{d^{2}}\) (b) \(\frac{G M m}{8 d^{2}}\) (c) \(\frac{G M m}{d^{2}}\left[1+\frac{1}{8\left(1+\frac{R}{2 d}\right)}\right]\) (d) \(\frac{G M m}{d^{2}}\left[1-\frac{1}{8\left(1-\frac{R}{2 d}\right)^{2}}\right]\)

As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from \(C\) to \(B\), moment of inertia increases. Therefore, angular velocity decreases. If, \(M=\) mass of platform \(R=\) radius of platform \(m=\) mass of tortoise moving along the chord \(A B\) \(\mathrm{a}=\) perpendicular distance of \(O\) from \(A B\). Initial angular momentum, \(l_{1}=m R^{2}+\frac{M R^{2}}{2}\) At any time \(t\), let the tortoise reach \(D\) moving with velocity \(\underline{v}\). \(\therefore\) \(A D=v t\) As $$\begin{aligned} &A C=\sqrt{R^{2}-a^{2}} \\ &D C=A C-A D=\left(\sqrt{R^{2}-a^{2}}-v t\right) \end{aligned}$$ \(\therefore\) $$O D=r=a^{2}+\left[\sqrt{R^{2}-a^{2}}-v t\right]^{2}$$ Angular momentum at time \(t\) $$ I_{2}=m r^{2}+\frac{M R^{2}}{2} $$ As angular momentum is conserved \(\therefore\) $$l_{1} \omega_{0}=I_{2} \omega(t)$$ This shows that variation of \(\omega(t)\) with time is non-linear.

The value of \(g\) on the earth's surface is \(980 \mathrm{cms}^{-2}\). Its value at a height of \(64 \mathrm{~km}\) from the earth's surface is (Radius of the earth \(R=6400 \mathrm{~km}\) ) (a) \(960.40 \mathrm{cms}^{-2}\) (b) \(984.90 \mathrm{cms}^{-2}\) (c) \(982.45 \mathrm{cms}^{-2}\) (d) \(977.55 \mathrm{cms}^{-2}\)

A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clock run at the same rate on earth. On a planet having the same density as earth but twice the radius, (a) \(S\) will run faster than \(P\) (b) \(P\) will run faster than \(S\) (c) both will run at the same rate as on the earth (d) both will run at the same rate which will be different from that on the earth

If the value of \(g\) acceleration due to gravity at earth surface is \(10 \mathrm{~ms}^{-2}\), its value in \(\mathrm{ms}^{-2}\) at the centre of the earth, which is assumed to be a sphere of radius \(R\) metre and uniform mass density is (a) 5 (b) \(10 / R\) (c) \(10 / 2 R\) (d) zero
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