A thin metal rod of mass1.3 kg and length0.4 m is at rest in outer space, near a space station (Figure 11.99). A tiny meteorite with mass 0.06 kg travelling at a high speed of strikes the rod a distance 0.2 m from the center and bounces off with speed 60 m/s as shown in the diagram. The magnitudes of initial and final angles to thex-axis of the small mass’s velocity are θi=26° and θf=82°.(a). Afterward, what is the velocity of the center of the rod? (b) Afterward, what is the angular velocity ω of the rod? (c) What is the increase in internal energy of the objects?

Short Answer

Expert verified

The velocity of the center of mass of the rod along horizontal is 8.68m/s.

The magnitude of the center of mass of the rod is 8.78 m/s.

The angular velocity of the rod is 130 rad/s.

The increase in internal kinetic energy of the system is 803J.

Step by step solution

01

Definition of law of conservation of momentum.

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects.

Use the law of conservation of momentum and the formula for the moment of inertia of the rod. The moment of inertia of the rod with respect to its center is as follows:

I=121ML2

Here, Mis the mass of the rod andLis the length of the rod.

The motion of the meteorite is shown in the following figure:

In the figure, mis the mass of the meteorite, dis the distance from the center of the rod to the point of application of velocity, Lis the length of the rod, andMis the mass of the rod.

02

Find the velocity of the center of mass of the rod.

  1. Calculate the center of mass of the rod after the collision as follows:

Apply the law of conservation of momentum along the horizontal as follows:

mvicosθi=-mvfcosθf+MVx,cm

HereVx,cmis the center of mass of the rod alongx-direction.

Rearrange the equation for the velocity of the center of mass of the rod.

Vx,cm=mvicosθi+mvfcosθfM

Substitute200m/sforvi,60m/sforvf,0.06kgfor m,1.3kgforM,26°for θi,and 82°forθf in the equation Vx,cm=mvicosθi+mvfcosθfMand solve for Vx,cm.

Vcm=(0.06kg)(200m/s)cos26°+(0.06kg)(60m/s)cos82°1.3kg

=8.68m/s

Therefore, the velocity of the center of mass of the rod along horizontal is 8.68m/s.

03

Find the magnitude of the center of mass of the rod.

Apply the law of conservation of momentum along the vertical as follows:

mvisinθi=mvf+MVfsinθf+MVy,cm

HereVy,cmis the center of mass of the rod alongy-direction.

Rearrange the equation for the velocity of the center of mass of the rod.

Vy,cm=mvisinθi-mvfsinθfM

Substitute 200m/sforvi,60m/s forvf,0.06kgform,1.3kgforM,26°for θi,and82°forθfin the equation Vy,cm=mvisinθi-mvfsinθfMand solve for Vy,cm

Vy,cm=(0.06kg)(200m/s)sin26°-(0.06kg)(60m/s)sin82°1.3kg

=1.3m/s

Therefore, the velocity of the center of mass of the rod along vertical is 1.3m/s.

Therefore, the center of mass of the rod is8.68m/s,1.3m/s,0m/s

The magnitude of center of mass of the rod is as follows:

Vcm=(8.68m/s)2+(1.3m/s)2+(0m/s)2=8.78m/s

Therefore, the magnitude of the center of mass of the rod is 8.78m/s.

04

Find the angular velocity of the rod.

b. Apply law of conservation of angular momentum.

Li=Lf

(mvicosθi)d=-(mvfcosθf)d+Iω(mvicosθi)d=-(mvfcosθf)d+Iω

HereIis the moment of inertia of the rod and is the angular velocity.

Substitute112ML2for(mvicosθi)d=-(mvfcosθf)d+Iωin the equation

mvicosθid=-mdvfcosθf+112ML2ωmvicosθid=-mdvfcosθf+112ML2ω

Rearrange the equation for the angular velocity.

ω=md(vicosθi+vfcosθf)112ML2

Substitute 0.4mforL,0.2mford,200m/sfor vi,60m/sforvf,0.06kgform,1.3kgfor M,26°for θi,and 82°for θf.

ω=(0.06kg)(0.2m)((200m/s)cos26°+(60m/s)cos82°)112(1.3kg)(0.4m)2

ω=(0.06kg)(0.2m)((200m/s)cos26°+(60m/s)cos82°)112(1.3kg)(0.4m)2

=130rad/s

Therefore, the angular velocity of the rod is130rad/s.

05

Find the internal kinetic energy of the system.

(c) Calculate the kinetic energy is of the system before collision as follows:

Initially the rod is in rest. So, initial kinetic of the rod is zero.

Ki=12mvi2

Calculate the kinetic energy is of the system before collision as follows:

Kf=12mvf2+12Iω2+12MVcm2

Calculate the increase in internal energy as follows:

The expression for the change in internal energy is as follows:

ΔE=Ki-KfΔE=12mvi2-12mvf2-12Iω2-12MVcm2

Substitute 112ML2for in the equation ΔE=12mvi2-12mvf2-12Iω2-12MVcm2and solve for ΔE.

ΔE=12m(vi2-vf2)-12112ML2ω2-12MVcm2

Substitute 0.4mforL,200m/sforv1,60m/sforvf,0.06kgfor m,1.3kgfor M,8.78m/sforVcm,and 130rad/s forωin the equation

ΔE=12m(vi2-vf2)-12112ML2ω2-12MVcm2and solve forΔE.

ΔE=0.06kg2(200m/s2-(60m/s)2-12112(1.3kg)(0.4m)2(130rad/s)2-12(1.3kg)(8.68m/s2+(1.3m/s)2

=895.46J895.5J

Therefore, the increase in internal kinetic energy of the system is 803J.

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Most popular questions from this chapter

A sick of length Land mass Mhangs from a low-friction axle (Figure 11.90). A bullet of mass mtravelling at a high speedstrikes vnear the bottom of the stick and quickly buries itself in the stick.

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