Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 33AP

Review. A thin, uniform, rectangular signboard hangs vertically above the door of a shop. The sign is hinged to a stationary horizontal rod along its top edge. The mass of the sign is 2.40 kg, and its vertical dimension is 50.0 cm. The sign is swinging without friction, so it is a tempting target for children armed with snowballs. The maximum angular displacement of the sign is 25.0° on both sides of the vertical. At a moment when the sign is vertical and moving to the left, a snowball of mass 400 g, traveling horizontally with a velocity of 160 cm/s to the right, strikes perpendicularly at the lower edge of the sign and sticks there. (a) Calculate the angular speed of the sign immediately before the impact. (b) Calculate its angular speed immediately after the impact. (c) The spattered sign will swing up through what maximum angle?

(a)

Expert Solution
Check Mark
To determine

The angular speed of the sign immediate before the collision.

Answer to Problem 33AP

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

Explanation of Solution

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, Chapter 11, Problem 33AP

Figure (I)

The height of the sign is calculated as,

    h=12L12Lcosθ=12L(1cosθ)

Here, L is the vertical length of the sign and θ is the angular displacement.

The formula for the conservation of the energy is,

    Mgh=Iω22ω2=2MghI        (I)

Here, M is the mass of sign, g is the acceleration due to gravity, I is the moment of inertia and 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)

Expert Solution
Check Mark
To determine

The angular speed of the sign immediate after the collision.

Answer to Problem 33AP

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

Explanation of Solution

The formula for the conservation of the angular momentum is,

    Ifωf=IiωimvL        (II)

Here, m is the mass of snowball, v is the velocity of snowball and 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)

Expert Solution
Check Mark
To determine

The maximum angle.

Answer to Problem 33AP

The maximum angle is 5.58°.

Explanation of Solution

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

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

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