EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.
EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.
14th Edition
ISBN: 9780133976588
Author: HIBBELER
Publisher: PEARSON CO
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Chapter 18.4, Problem 1PP

Determine the kinetic energy of the 100-kg object.

Chapter 18.4, Problem 1PP, Determine the kinetic energy of the 100-kg object.

a)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is 900J .

Explanation of Solution

Given:

The mass of disk is 100kg .

The angular velocity of the disk is ω=3rad/s .

The radius of the disk is 2m .

Draw the free body diagram of the rod as shown in Figure (1a).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  1

Refer Figure (1a).

Write the formula for mass moment of inertia (I) of the disk about it is center.

I=12mr2

Here, m is the mass and r is the radius of the disk.

Write the formula for kinetic energy (T)

(Rotation about fixed axis).

T=12Iω2

Substitute 12mr2 for I .

T=12(12mr2)ω (I)

Here, ω is the angular velocity.

Conclusion:

Calculate the kinetic energy of the disk.

Substitute 100kg for m , 2m for r and 3rad/s for ω in Equation (I).

T=12[12(100kg)(2m)2](3rad/s)2=12(200)(9)=900J

Thus, the kinetic energy of the disk is 900J .

b)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is 800J .

Explanation of Solution

Given:

The mass of rod is 100kg .

The angular velocity of the rod is ω=2rad/s .

The length of the rod is 6m .

Draw the free body diagram of the rod as shown in Figure (1b).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  2

Refer Figure (1b).

Write the formula for mass moment of inertia (IO) of the rod about the point O .

IO=IG+mlOG2                      (I)

Here, IG is the moment of inertial about the centroid G and IG=112ml2 .

Substitute 112ml2 for IG in Equation (I).

IO=112ml2+mlOG2

Here, IO is the mass moment of inertia about point O , m is the mass, l is the total length of the rod and lOG , distance between the point OandG .

Write the formula for kinetic energy (T) .

T=12IOω2

Substitute 112ml2+mlOG2 for IO .

T=12(112ml2+mlOG2)ω2 (I)

Here, ω is the angular velocity.

Conclusion:

Refer Figure (1b).

Calculate the kinetic energy of the rod.

Substitute 100kg for m , 6m for l , 1m for lOG and 2rad/s for ω in Equation (I).

T=12[112(100kg)(6m)2+(100kg)(1m)2](2rad/s)2=12(400)(4)=800J

Thus, the kinetic energy of the rod is 800J .

c)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is 1200J .

Explanation of Solution

Given:

The mass of disk is 100kg .

The angular velocity of the disk is ω=3rad/s .

The radius of the disk is 2m .

Draw the free body diagram of the disk as shown in Figure (1c).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  3

Refer Figure (1c).

Write the formula for mass moment of inertia (I) of the disk about it is center.

I=12mr2

Here, m is the mass and r is the radius of the disk.

Write the formula for kinetic energy (T) .

T=12mv2+12Iω2

Here, v=rω .

Substitute rω for v and 12mr2 for I .

T=12m(rω)2+12(12mr2)ω2 (I)

Here, m is the mass, r is radius, I is the mass moment of inertia and ω is the angular velocity.

Conclusion:

Refer Figure (1c).

Calculate the kinetic energy of the disk.

Substitute 100kg for m , 2m for r and 3rad/s for ω in Equation (I).

T=[12(100kg)(2m×2rad/s)2]+12[12(100kg)(2m)2](2rad/s)2=800+400=1200J

Thus, the kinetic energy of the disk is 1200J .

d)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the rod is 600J .

Explanation of Solution

Given:

The mass of rod is 100kg .

The angular velocity of the rod is ω=2rad/s .

The length of the rod is 3m .

Draw the free body diagram of the rod as shown in Figure (1d).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  4

Refer Figure (1d).

Write the formula for mass moment of inertia (IO) of the rod having the rotation axis at one end.

IO=13ml2

Here, m is the mass and l is the length of the rod about mass center O.

Write the formula for kinetic energy (T)

(Rotation about fixed axis).

T=12IOω2

Substitute 13ml2 for IO .

T=12(13ml2)ω2 (I)

Here, IO is the mass moment of inertia and ω is the angular velocity.

Conclusion:

Refer Figure (1d).

Calculate the kinetic energy of the rod.

Substitute 100kg for m , 3m for l and 2rad/s for ω in Equation (I).

T=12[13(100kg)(3m)2](2rad/s)2=12(300)(4)=600J

Thus, the kinetic energy of the rod is 600J .

e)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is 4800J .

Explanation of Solution

Given:

The mass of disk is 100kg .

The angular velocity of the disk is ω=4rad/s .

The radius of the disk is 2m .

Draw the free body diagram of the disk as shown in Figure (1e).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  5

Refer Figure (1e).

Write the formula for mass moment of inertia (I) of the disk about it is center.

I=12mr2

Here, m is the mass and r is the radius of the disk.

Write the formula for kinetic energy (T) .

T=12mv2+12Iω2

Here, v=rω .

Substitute rω for v and 12mr2 for I .

T=12m(rω)2+12(12mr2)ω2 (I)

Here, m is the mass, r is radius, I is the mass moment of inertia and ω is the angular velocity.

Conclusion:

Refer Figure (1e).

Calculate the kinetic energy of the disk.

Substitute 100kg for m , 2m for r and 4rad/s for ω in Equation (I).

T=[12(100kg)(2m×4rad/s)2]+12[12(100kg)(2m)2](4rad/s)2=3200+1600=4800J

Thus, the kinetic energy of the disk is 4800J .

f)

Expert Solution
Check Mark
To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is 3200J .

Explanation of Solution

Given:

The mass of rod is 100kg .

The angular velocity is ω=4rad/s .

Draw the free body diagram of the object as shown in Figure (1f).

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC., Chapter 18.4, Problem 1PP , additional homework tip  6

Refer Figure (1f).

Here, the ends of the rod are connected to two rods of same length. Hence the rod travels in circular motion.

Consider as the mass travels in a radius of r=2m .

Write the formula for kinetic energy (T)

(Rotation about fixed axis).

T=12mv2

Here, v=rω .

Substitute rω for v .

T=12m(rω)2 (I)

Here, m is the mass, r is radius and ω is the angular velocity.

Conclusion:

Refer Figure (1f).

Here r=2m .

Calculate the kinetic energy of the disk.

Substitute 100kg for m , 2m for r and 4rad/s for ω in Equation (I).

T=12(100kg)[(4rad/s)(2m)]2=50(64)=3200J

Thus, the kinetic energy of the disk is 3200J .

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Chapter 18 Solutions

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.

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