Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 16, Problem 112P
To determine

The angular speed of the dipole when it reaches θ=0, if it was released from rest at θ=90.0°.

Expert Solution & Answer
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Answer to Problem 112P

The angular speed of the dipole when it reaches θ=0, if it was released from rest at θ=90.0°, is 20rad/s.

Explanation of Solution

Write the expression for the torque acting on the dipole.

τ=qEdsinθ (I)

Here, τ is the torque on the dipole, q is the charge, E is the electric field acting, d is dipole length and θ is the angle the dipole makes with electric field

Write the expression for the angular acceleration.

α=τI

Here, α is the angular acceleration and I is the moment of inertia of the dipole

Put equation (I) in the above equation.

α=(qEdI)sinθ (II)

Since the expression for the angular acceleration contains sinθ , it is a variable function. This implies Newton’s constant acceleration equations of motion cannot be applied here.

The best way to calculate angular speed is to use right average angular acceleration. This average acceleration must lie between qEd/I and 0 within the range of θ=0 to 0=π2.

Substitute π2 for θ to get αmax and 0 for θ to get αmin in equation (II).

αmax=(qEdI)sinπ2=qEdI

αmin=(qEdI)sin0=0

Here, αmin is the minimum angular acceleration and αmax is the maximum angular acceleration.

The following figure gives the graph of sinθ for the interval θ=0to θ=π2.

Physics, Chapter 16, Problem 112P

Above figure indicates that average value of sinθ in this range is larger than one half of maximum value, because of the shape of the graph.

Consider the area of rectangle having height equal to right average acceleration and width π2.This area has same value as area below the graph of sinθ versus θ in the limit θ=0 to θ=π2.

From more advanced mathematics, area of the sine function is the equal to 1, when θ is in radians.

Write the expression for the area of rectangle.

sinθavg(π2)=1sinθavg=2π

Write the expression for the angular velocity.

ωf2ωi2=2α(θfθi) (III)

Here, ωf is the final angular velocity, ωi is the initial angular velocity, θf is the final angular position and θi is the initial angular position.

The dipole system consists of charges at end of uniform rod.

Write the expression for moment of inertia of dipole system.

I=mr2+(112)ML2+mr2 (IV)

Here, m is the mass of charge, r is the distance between center to end of rod, M is the mass of rod and L is the length of rod.

Conclusion:

Substitute 5.0g for m , 20.0g for M , 3.5cm for r and 7.0cm for L in equation (IV) to get I.

I=(5.0g×1kg1000g)(3.5cm×1m100cm)2+(112)(20.0g1kg1000g)(7.0cm×1m100cm)2+(5.0g×1kg1000g)(3.5cm×1m100cm)2=2.0×105kgm2

Substitute 0rad/s for ωi , qEdIsinθavg for α , π2 for θf and 0 for θi in equation(III) to get ωf.

ωf2(0rad/s)2=2×qEdIsinθavg(π20)ωf2=qπEdIsinθavg

Substitute 2π for sinθavg in above equation to get the final expression for ωf.

ωf2=qπEdI2πωf=2qEdI

Substitute 3.0μC for q , 2.0×104N/C for E , 7.0cm for d and 2.0×105kgm2 for I in above equation to get ωf.

ωf=2(3.0μC×1C106μC)(2.0×104N/C)(7.0cm×1m100cm)2.0×105kgm21kgm1Ns21rad1=20rad/s

Therefore, the angular speed of the dipole when it reaches θ=0, if it was released from rest at θ=90.0°, is 20rad/s.

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