Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Question
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Chapter 9, Problem 118P

(a)

To determine

To Calculate: The initial angular acceleration of the structure.

(a)

Expert Solution
Check Mark

Answer to Problem 118P

  0.3rad/s2

Explanation of Solution

Given information:

Height of each vertical beam =12.0ft

Width of each vertical beam =2.0ft

Length horizontal cross-member =6.0ft

The mass of the vertical beam =350 kg

The mass of the horizontal beam =280 kg

Formula Used:

From Newton’s second law of motion

  F=ma

Where, F is the net force, m is the mass and a is the acceleration.

Calculation:

  Physics for Scientists and Engineers, Vol. 1, Chapter 9, Problem 118P , additional homework tip  1

Here, one of the structures initially in its upright position. But later, it is about to strike the floor. The moments about the axis of rotation (a line through point P) can be considered.

Use the parallel-axis theorem to find the moments of inertia of the two parts of this composite structure. Let the numeral 1 denote the vertical member and the numeral 2 the horizontal member.

Apply Newton’s second law of motion in rotational form to the structure to express its angular acceleration in terms of the net torque causing it to fall and its moment of inertia with respect to point P.

Taking clockwise rotation to be positive, use τ=IPα .

  m2g(l22)m1g(w2)=Ipα

Here,

The mass of the vertical beam = m1

mass of the horizontal beam = m2

Length of vertical beam = l2

Width of the beam = w

Acceleration of the beam = α

Gravitational acceleration = g

  α=m2gl2m1gw2Ip

Because IP = I1P + I2P

  α=g(m2l2m1w)2(I1p+I2p)

  l1=10.0ft×1m3.281ftl1=3.05m

  l1=6.0ft×1m3.281ftl1=1.83m

  w=2.0ft×1m3.281ftl1=0.61m

  I1p=13m1l12+m1(w2)2I1p=m1(13l12+14w2)

  I1p=(350kg)[13(3.05m)2+14(0.610m)2]I1p=1.12×103kgm2

  I2p=I2,cm+m2d2

  d2=(l1+12w)2+(12l2w)2d=(l1+12w)2+(12l2w)2

  d=[3.05m+(0.610m)]2+[12(1.83m)0.610m]2=3.67m

  I2,cm=112m2l22

  I2p=112m2l22+m2d2I2p=m2(112l22+d2)

  I2p=(280kg)[112(1.83m)2+(3.67m)2]I2p=3.85×103kgm2

  α=(9.81ms-2)[(280kg)(1.83m)(350kg)(0.61m)]2(1.12+3.85)×103kgm2α=0.3rad/s2

Conclusion:

The initial angular acceleration of the structure is 0.3rad/s2 .

(b)

To determine

ToCalculate: The magnitude of the initial linear acceleration of the right end of the horizontal beam.

(b)

Expert Solution
Check Mark

Answer to Problem 118P

  1.16ms-2

Explanation of Solution

Given information:

Height of each vertical beam =12.0ft

Width of each vertical beam =2.0ft

Length horizontal cross-member =6.0ft

The mass of the vertical beam =350 kg

The mass of the horizontal beam =280 kg

The initial angular acceleration of the structure is 0.3rad/s2 .

Formula used:

From Newton’s second law of motion

  F=ma

Where, F is the net force, m is the mass and a is the acceleration.

Linear acceleration (a) in terms of angular acceleration ( α ) and radius (R) is:

  a=αR

Calculation:

  Physics for Scientists and Engineers, Vol. 1, Chapter 9, Problem 118P , additional homework tip  2

R2=(l1+w)+(l2w)

  R=(l1+w)2+(l2w)2R=(3.05m+0.61m)2+(1.83m-0.61m)2R=3.86m

  a=(0.3rad/s2)(3.86m)a=1.16ms-2

Conclusion:

The magnitude of the initial linear acceleration of the right end of the horizontal beam is 1.16ms-2 .

(c)

To determine

To explain: The horizontal component of the initial linear acceleration be at this same location.

(c)

Expert Solution
Check Mark

Answer to Problem 118P

  1.1ms-2

Explanation of Solution

Given information:

Height of each vertical beam =12.0ft

Width of each vertical beam =2.0ft

Length horizontal cross-member =6.0ft

The mass of the vertical beam =350 kg

The mass of the horizontal beam =280 kg

The magnitude of the initial linear acceleration of the right end of the horizontal beam is 1.16ms-2 .

Formula used:

Horizontal component of acceleration is ax=acosθ .

Calculation:

  ax=acosθax=al1+wR

  ax=(1.16ms-2)3.05m+0.610m3.86max=1.1ms-2

Conclusion:

The horizontal component of the initial linear acceleration be at this same location is 1.1ms-2 .

(d)

To determine

To Calculate: The beam’s rotational speed when they caught it.

(d)

Expert Solution
Check Mark

Answer to Problem 118P

  1.05rad/s

Explanation of Solution

Given information:

Height of each vertical beam =12.0ft

Width of each vertical beam =2.0ft

Length horizontal cross-member =6.0ft

The mass of the vertical beam =350 kg

The mass of the horizontal beam =280 kg

Formula used:

Rotational kinetic energy:

  K.Er=12Iω2

Where, I is the moment of inertia and ω is the angular speed.

Potential energy, U=mgh

Where, m is the mass, g is the acceleration due to gravity and h is the height.

Calculation:

By applying the conservation of mechanical energy to the beam:

  ΔK+ΔUs=KfKi+UfUi=0

  Ug=0

  KfUi=0

  12Ipωf2mg(Δh)=0

  Δh14l1

  12Ipωf214mgl1=0ωf=mgl12Ip

  ωf=(350kg)(9.81ms-2)(3.05m)2(1.12+3.85)×103kgm-2ωf=1.05rad/s

Conclusion:

The beam’s rotational speed when they caught it is 1.05rad/s .

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Chapter 9 Solutions

Physics for Scientists and Engineers, Vol. 1

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