Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 33AP

(a)

To determine

The angular speed of the sign immediate before the collision.

(a)

Expert Solution
Check Mark

Answer to Problem 33AP

The angular speed of the sign immediate before the collision is 2.35 rad/s .

Explanation of Solution

Given information:

The mass of sign is 2.40 kg , the vertical length of sign is 50 cm , the maximum angular displacement is 25° , the mass snowball is 400 g and the velocity of snowball is 160 cm/s .

Consider figure given below.

Physics for Scientists and Engineers with Modern Physics, Chapter 11, Problem 33AP

Figure (I)

The height of the sign is calculated as,

h=12L12Lcosθ=12L(1cosθ)

  • L is the vertical length of the sign.
  • θ is the angular displacement.

The formula for the conservation of the energy is,

Mgh=Iω22ω2=2MghI (I)

  • M is the mass of sign.
  • g is the acceleration due to gravity.
  • I is the moment of inertia.
  • ωi is the angular speed before the collision.

The formula to calculate initial moment of inertia is,

Ii=ML212+M(L2)2=ML212+ML24=ML23

Substitute 12L(1cosθ) for h and ML23 for Ii in equation (I) to find ωi .

ωi2=2Mg12L(1cosθ)ML23=3g(1cosθ)Lωi=3g(1cosθ)L

Substitute 9.8 m/s2 for g , 25° for θ and 50 cm for L in above equation to find ωi .

ωi=3(9.8 m/s2)(1cos(25°))50 cm(102 m1 cm)=2.35 rad/s

Conclusion:

Therefore, the angular speed of the sign immediate before the collision is 2.35 rad/s .

(b)

To determine

The angular speed of the sign immediate after the collision.

(b)

Expert Solution
Check Mark

Answer to Problem 33AP

The angular speed of the sign immediate after the collision is 0.498 rad/s .

Explanation of Solution

Given information:

The mass of sign is 2.40 kg , the vertical length of sign is 50 cm , the maximum angular displacement is 25° , the mass snowball is 400 g and the velocity of snowball is 160 cm/s .

The formula for the conservation of the angular momentum is,

Ifωf=IiωimvL (II)

  • m is the mass of snowball.
  • v is the velocity of snowball.
  • ωf is the angular speed after the collision.

The formula to calculate final moment of inertia is,

If=(ML23+mL2)

Substitute (ML23+mL2) for If and ML23 for Ii in equation (II) to find ωf .

(ML23+mL2)ωf=(ML23)ωimvLωf=(ML23)ωimvL(ML23+mL2)=(ML23)ωimvLL2(M3+m)

Substitute 2.40 kg for M , 400 g for m , 2.35 rad/s for ωi , 160 cm/s for v and 50 cm for L in above equation to find ω .

ωf=((2.40 kg)(50 cm)23)(2.35 rad/s)(400 g)(160 cm/s)(50 cm)(50 cm)2((2.40 kg)3+(400 g))=(2000 kgcm2)(2.35 rad/s)(400 g(103 kg1 g))(160 cm/s)(50 cm)(50 cm)2((2.40 kg)3+(400 g(103 kg1 g)))=(2000 kgcm2(104 m21 cm2))(2.35 rad/s)(3200 kgcm2(104 m21 cm2)/s)(50 cm(102 m1 cm))2(1.2 kg)=0.498 rad/s

Conclusion:

Therefore, the angular speed of the sign immediate after the collision is 0.498 rad/s .

(c)

To determine

The maximum angle.

(c)

Expert Solution
Check Mark

Answer to Problem 33AP

maximum angle is 5.58° .

Explanation of Solution

Section 1:

To determine: The distance of the centre of mass from the axis of rotation.

Answer: The distance of the centre of mass from the axis of rotation is 0.28 m .

Given information:

The mass of sign is 2.40 kg , the vertical length of sign is 50 cm , the maximum angular displacement is 25° , the mass snowball is 400 g and the velocity of snowball is 160 cm/s .

The formula distance of the centre of mass from the axis of rotation is,

hCM=m1y1+m2y2m1+m2

Substitute 2.40 kg for m1 , 400 g for m2 , 50 cm2 for y1 and 50 cm for y2 in above equation.

hCM=(2.40 kg)((50 cm2)(102 m1 cm))+(400 g(103 kg1 g))(50 cm)(102 m1 cm)(2.40 kg)+(400 g(103 kg1 g))=(2.40 kg)(0.250 m)+(0.400 kg)(0.500 m)(2.40 kg)+(0.400 kg)=0.28 m

Section 2:

To determine: The maximum angle.

Answer: The maximum angle is 5.58° .

Given information:

The mass of sign is 2.40 kg , the vertical length of sign is 50 cm , the maximum angular displacement is 25° , the mass snowball is 400 g and the velocity of snowball is 160 cm/s .

The conservation of mechanical energy is,

(M+m)ghCM(1cosθ)=12(ML23+mL2)ωf2(1cosθ)=12(ML23+mL2)ωf2(M+m)ghCM

Rearrange the above expression for cosθ .

cosθ=[1(M3+m)L2ωf22(M+m)ghCM]

Substitute 2.40 kg for M , 400 g for m , 0.498 rad/s for ωf , 50 cm for L , 9.8 m/s2 for g and 0.28 m for hCM in above equation.

cosθ=[1((2.40 kg)3+(400 g(103 kg1 g)))(50 cm(102 m1 cm))2(0.498 rad/s)22((2.40 kg)+(400 g(103 kg1 g)))(9.8 m/s2)(0.28 m)]θ=cos1[1(0.80 kg+(0.400 kg))(0.500 m)2(0.498 rad/s)22((2.40 kg)+(0.400 kg))(9.8 m/s2)(0.28 m)]=5.58°

Conclusion:

Therefore, the maximum angle is 5.58° .

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Chapter 11 Solutions

Physics for Scientists and Engineers with Modern Physics

Ch. 11 - A particle is located at a point described by the...Ch. 11 - A 1.50-kg particle moves in the xy plane with a...Ch. 11 - Prob. 9PCh. 11 - Heading straight toward the summit of Pikes Peak,...Ch. 11 - Review. A projectile of mass m is launched with an...Ch. 11 - Prob. 12PCh. 11 - A particle of mass m moves in a circle of radius R...Ch. 11 - A 5.00-kg particle starts from the origin at time...Ch. 11 - A ball having mass m is fastened at the end of a...Ch. 11 - Prob. 16PCh. 11 - A uniform solid disk of mass m = 3.00 kg and...Ch. 11 - Show that the kinetic energy of an object rotating...Ch. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - A 60.0-kg woman stands at the western rim of a...Ch. 11 - Prob. 24PCh. 11 - A uniform cylindrical turntable of radius 1.90 m...Ch. 11 - Prob. 26PCh. 11 - A wooden block of mass M resting on a...Ch. 11 - Prob. 28PCh. 11 - A wad of sticky clay with mass m and velocity vi...Ch. 11 - A 0.005 00-kg bullet traveling horizontally with a...Ch. 11 - The angular momentum vector of a precessing...Ch. 11 - A light rope passes over a light, frictionless...Ch. 11 - Prob. 33APCh. 11 - Prob. 34APCh. 11 - We have all complained that there arent enough...Ch. 11 - Prob. 36APCh. 11 - A rigid, massless rod has three particles with...Ch. 11 - Prob. 38APCh. 11 - Two astronauts (Fig. P11.39), each having a mass...Ch. 11 - Two astronauts (Fig. P11.39), each having a mass...Ch. 11 - Native people throughout North and South America...Ch. 11 - Two children are playing on stools at a restaurant...Ch. 11 - You are attending a county fair with your friend...Ch. 11 - Prob. 44APCh. 11 - Global warming is a cause for concern because even...Ch. 11 - The puck in Figure P11.46 has a mass of 0.120 kg....Ch. 11 - Prob. 47APCh. 11 - A solid cube of wood of side 2a and mass M is...Ch. 11 - Prob. 49CPCh. 11 - Prob. 50CP
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