PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 8, Problem 4P

(a)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about x axis.

(a)

Expert Solution
Check Mark

Answer to Problem 4P

The rotational inertia of the system of point particles about x axis is 13,000gcm2.

Explanation of Solution

Write the expression for the rotational inertia about x axis.

Ix=i=ACmiri2

Here, Ix is the moment of inertia about x axis, mi is the mass of ith particle and ri is distance of ith particle.

Expand above summation.

Ix=mArA2+mBrB2+mCrC2

Here, mA is the mass of particle at A , mB is the mass of particle at B , mC is the mass of particle at C , rA is the perpendicular distance of point A from x axis, rB is the perpendicular distance of point B from x axis and rC is the perpendicular distance of point C from x axis.

Conclusion:

Substitute 200g for mA , 300g for mA, 500g for mC, 5.0cm for rA , 0cm for rB and 4.0cm for rC in above equation to get Ix.

Ix=(200g)(5.0cm)2+(300g)(0cm)2+(500g)(4.0cm)2=13,000gcm2

Therefore, the rotational inertia of the system of point particles about x axis is 13,000gcm2.

(b)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about y axis.

(b)

Expert Solution
Check Mark

Answer to Problem 4P

The rotational inertia of the system of point particles about y axis is 25,000gcm2.

Explanation of Solution

Write the expression for the rotational inertia about y axis.

Iy=i=ACmiri2

Here, Ix is the moment of inertia about y axis, mi is the mass of ith particle and ri is distance of ith particle.

Expand above summation.

Iy=mArA2+mBrB2+mCrC2

Here, mA is the mass of particle at A , mB is the mass of particle at B , mC is the mass of particle at C , rA is the perpendicular distance of point A from x axis, rB is the perpendicular distance of point B from x axis and rC is the perpendicular distance of point C from y axis.

Conclusion:

Substitute 200g for mA , 300g for mA, 500g for mC, 3.0cm for rA , 6.0cm for rB and 5.0cm for rC in above equation to get Iy.

Iy=(200g)(3.0cm)2+(300g)(6.0cm)2+(500g)(5.0cm)2=25,000gcm2

Therefore, the rotational inertia of the system of point particles about y axis is 25,000gcm2.

(c)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about z axis.

(c)

Expert Solution
Check Mark

Answer to Problem 4P

The rotational inertia of the system of point particles about z axis is 38,000gcm2.

Explanation of Solution

Write the expression for the rotational inertia about z axis.

Iz=i=ACmiri2

Here, Iz is the moment of inertia about z axis, mi is the mass of ith particle and ri is distance of ith particle.

Expand above summation.

Iz=mArA2+mBrB2+mCrC2

Here, mA is the mass of particle at A , mB is the mass of particle at B , mC is the mass of particle at C , rA is the perpendicular distance of point A from x axis, rB is the perpendicular distance of point B from x axis and rC is the perpendicular distance of point C from z axis.

Conclusion:

Substitute 200g for mA , 300g for mA, 500g for mC, (3.0cm)2+(5.0cm)2 for rA , 6.0cm for rB and (5.0cm)2+(4.0cm)2 for rC in above equation to get Iz.

Iz=(200g)[(3.0cm)2+(5.0cm)2]+(300g)(6.0cm)2+(500g)[(5.0cm)2+(4.0cm)2]=38,000gcm2

Therefore, the rotational inertia of the system of point particles about z axis is 38,000gcm2.

(d)

To determine

The x and y coordinates of center of mass of system.

(d)

Expert Solution
Check Mark

Answer to Problem 4P

The x and y coordinates of center of mass of system is (x,y)=(1.3cm,1.0cm).

Explanation of Solution

Write the expression for the x coordinate of center of mass of system.

xCM=mAxA+mBxB+mCxCmA+mB+mC (I)

Here, xCM is the x coordinate of center of mass system, xA is the x coordinate of center of mass, xB is the x coordinate of center of mass of B and xC is the x coordinate of center of mass of

C.

Write the expression for the y coordinate of center of mass of system.

yCM=mAyA+mByB+mCyCmA+mB+mC (II)

Here, yCM is the y coordinate of center of mass system, yA is the y coordinate of center of mass, yB is the y coordinate of center of mass of B and yC is the y coordinate of center of mass of

C.

Conclusion:

Substitute 200g for mA , 300g for mA, 500g for mC, 3.0cm for xA , 6.0cm for xB and 5.0cm for xC in equation (I) to get xCM.

xCM=(200g)(3.0cm)+(300g)(6.0cm)+(500g)(5.0cm)200g+300g+500g=1.3cm

Substitute 200g for mA , 300g for mA, 500g for mC, 5.0cm for xA , 0cm for xB and 4.0cm for xC in equation (II) to get yCM.

xCM=(200g)(5.0cm)+(300g)(0cm)+(500g)(4.0cm)200g+300g+500g=1.0cm

Therefore, the x and y coordinates of center of mass of system is (x,y)=(1.3cm,1.0cm).

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Chapter 8 Solutions

PHYSICS

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