/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Asteroid Collision! Suppose that... [FREE SOLUTION] | 91Ó°ÊÓ

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Asteroid Collision! Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass \(M,\) for the day to become 25.0\(\%\) longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Short Answer

Expert verified
The asteroid must have a mass of \( \frac{M}{5} \), one-fifth of Earth's mass.

Step by step solution

01

Understand the Concepts

To solve this problem, we need to understand angular momentum and rotational inertia. When the asteroid collides with Earth and remains embedded, angular momentum is conserved.
02

Define the Problem

We want to find the mass of the asteroid that makes Earth's day 25% longer. Increase in day length implies an increase in Earth's period of rotation.
03

Calculate Earth's Initial Angular Momentum

Calculate Earth's initial angular momentum using the formula: \[ L_i = I_i \omega_i \]where \( I_i = \frac{2}{5}MR_e^2 \) is Earth's initial moment of inertia (assuming a solid sphere), and \( \omega_i = \frac{2\pi}{T} \) is its initial angular velocity, with \( T = 24 \) hours.
04

Calculate Earth's Final Angular Momentum

The final angular speed \( \omega_f \) is given as 25% longer day, i.e., \( T_f = 1.25T \). Thus, \( \omega_f = \frac{2\pi}{1.25T} \). Earth's final moment of inertia after the collision is \[ I_f = I_i + mR_e^2 \] where \( m \) is the asteroid's mass.
05

Set Up Angular Momentum Conservation Equation

Since angular momentum is conserved, \[ L_i = L_f \]Hence,\[ I_i \omega_i = I_f \omega_f \]
06

Solve for the Mass of the Asteroid

Substitute known values into the equation from Step 5:\[ \frac{2}{5}MR_e^2 \omega_i = (\frac{2}{5}MR_e^2 + mR_e^2)\omega_f \]Cancelling \( R_e^2 \) and simplifying gives:\[ \frac{2}{5}M \frac{2\pi}{T} = (\frac{2}{5}M + m)\frac{2\pi}{1.25T} \]Solve the equation for \( m \):\[ m = \frac{2}{5}M \left( \frac{1.25T}{T} - 1 \right) \left(\frac{1}{\frac{1.25T}{T}}\right) \]This simplifies to:\[ m = \frac{M}{5} \]
07

Conclusion

The mass of the asteroid required to increase the length of a day by 25% is \( \frac{M}{5} \), or one-fifth of the Earth's mass.

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, refers to the resistance an object experiences against changes in its rotational state. Just like mass is the measure of an object's resistance to linear motion changes, rotational inertia plays a similar role in rotational motion. It tells us how much torque is required to change an object's angular velocity.

- Rotational inertia depends on both the mass of the object and how that mass is distributed relative to the axis of rotation.- For a solid sphere, the formula for rotational inertia is given by: \( I = \frac{2}{5} M R^2 \), where \( M \) is the mass and \( R \) is the radius. This specific formula was used in the given problem to understand Earth's initial resistance to rotational speed change.

Overall, rotational inertia is a key concept, especially when calculating how a system, like Earth, reacts to an external force like an asteroid impact.
Moment of Inertia
The moment of inertia is a specific measure of rotational inertia, symbolized as \( I \). It reflects how hard it is to change an object's rotational speed around a particular axis. The distribution of the mass relative to the axis also affects this measurement.

- The further the mass is from the rotation axis, the higher the moment of inertia.- Different shapes have different moment of inertia formulas: A solid sphere uses \( I = \frac{2}{5} M R^2 \), while a solid cylinder uses \( I = \frac{1}{2} M R^2 \).

For Earth in this exercise, we considered it as a solid sphere, using \( I = \frac{2}{5} M R^2 \). This helped in understanding both its initial and final states of rotation post-collision with the asteroid. The change in Earth's moment of inertia after the collision impacts its angular velocity, hence affecting how long a day lasts.
Angular Velocity
Angular velocity represents how fast an object rotates or spins. It's defined as the angle an object rotates through a unit of time, typically measured in radians per second.

- It's directly proportional to the rotation angle and inversely proportional to the time taken.- For a day 25% longer than usual, we calculated the new angular velocity using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the rotation period.

Changes in angular velocity affect the rotational dynamics of an object. For Earth, the impact of an asteroid halts some angular momentum, altering its speed and leading to longer days. Understanding this concept is essential for predicting how external forces influence planetary motion.
Rotational Motion Dynamics
Rotational motion dynamics studies how forces affect rotational motion. It includes concepts like torque, angular momentum, and rotational inertia to describe objects' rotational behavior.

- Angular momentum conservation is crucial in understanding rotational dynamics. It states that an object remains in rotational motion unless acted upon by an external force. - In the asteroid collision scenario, Earth's angular momentum is conserved: initial angular momentum equals final angular momentum. Applying this principle allowed us to determine the mass of the asteroid needed to increase the Earth's rotation period by 25%.

Understanding rotational motion dynamics helps us predict how a rotating object reacts to changes in external forces. When an asteroid impacts Earth, it alters the distribution of mass, increasing rotational inertia, and modifying the planet's rotational period.

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Most popular questions from this chapter

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . (a) If the disk rolls up a \(30.0^{\circ}\) ramp, how far along the ramp will it move before it stops? (b) Explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?(b)\) At the time \(3.00 s,\) find \((i)\) the angular momentum of the sphere and (ii) the net torque on the sphere.

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\) , connected by a light axle of radius b. A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

A 392 -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2}\) . Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J}\) . Calculate \(h\) .

One force acting on a machine part is \(\overrightarrow{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\boldsymbol{\imath}}+\) \((4.00 \mathrm{N}) \hat{\jmath}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\jmath} .\) (a) In a sketch, show \(\vec{r}, \vec{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

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