Chapter H - Problems

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Blinn College - Physics 2425 - Terry Honan

Problem H.1

A wheel rotates from rest to 12 radês in 3 s. Assume the angular acceleration is constant. (a) What is the magnitude of the wheel's angular acceleration?

(b) Through what angle (in radians) does the wheel rotate?

Solution to H.1 HaLw = w₀+ at ï 12=0+ a3 ï a =4 rad s HbLDq =1 2Hw0+ wLt= 1 2H0+12L3=18 rad

Problem H.2

(a) What is the angular velocity of the Earth in its orbit about the Sun. (b) What is the angular velocity of the Moon in its orbit about the Earth.

Solution to H.2

A constant angular velocity is related to the period T by w =2p T .

HaLT=1 year=365.24µ24µ3600=3.1557µ107s ï w =1.99µ10-7 rad s HbLT=27.32 days=27.32µ24µ3600=2.36 05µ106s ï w =2.66µ10-6 rad

s

Problem H.3

A grinding wheel rotating at 100 revêmin is turned off. It slows with a constant angular acceleration of 2 radês. (a) How long does it take the wheel to stop?

(b) Through what angle, in radians, does it turn while slowing?

Solution to H.3 w =0 and w₀=100 rev minµ 2prad rev µ 1 min 60 s =10.472 rad s HaLw = w₀+ at ï 0=10.472-2t ï 5.24 s HbLw2- w₀2=2a Dq ï Dq =0-10.4722 2µ2 =27.4 rad

Problem H.4

A car races around a circular track of radius 250 m at a constant speed of 45mês. (a) What is its angular velocity?

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Solution to H.4 R=250 m and v=45m HaLv=rw ï w = v R=0.18 rad s HbLac= w2r=v 2 R =8.1 m

s2 The direction of the acceleration is centripetal, meaning that is directed toward the center.

Problem H.5

By rotating through 1.25 rev of a 1 m radius arc, a discus thrower uniformly accelerates a discus from rest to 25mês. (a) What is the final angular velocity of the discus?

(b) What is its angular acceleration? (c) What is the total time of acceleration?

Solution to H.5 HaLv=rw ï w =v r= 25 1 =25 rad s HbLw2-w₀2=2a Dq ï 252-0=2aH2p µ1.25L ï a =39.8rad s HcLDq =1 2Hw0+ wLt ï 2p µ1.25= 1 2H0+25Lt ï t=0.628 s

Problem H.6

Assume that a baseball is a uniform sphere of radius 3.80 cm. If it moves at a speed of 38mês (the speed of the center of mass) and rotates at 125 radês then what is the ratio of the rotational kinetic energy to the translational kinetic energy.

Solution to H.6

For a uniform solid sphere I= 2

5MR 2 Krot=1₂Iw2= 1₂ ₅2M R2w2= 1₅M R2w2 Krot K_trans= 1 5M R2w2 1 2M v2 =2R2w2 5v2

Insert the numbers: R=0.038m, v=38m

s andw =125 rad s giving K_rot K_trans=0.00625 .

Problem H.7

Three masses glued onto a light rigid sheet in the xy plane. The masses and positions are: 4 kg at H-3 m, 5 mL, 6 kg at H4 m, 0L and 8 kg at H0,-3 mL.

(a) Suppose this sheet is rotated about the y axis at an angular velocity of 12 radês. What is its total rotational kinetic energy? (b) Suppose it rotates about the z axis at 12 rad/s. What is its rotational kinetic energy?

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Solution to H.7

(a) For a discrete distribution of mass: I=⁄_imiri2 where ri is the perpendicular distance of the mass mi from the axis. If the

axis is the y axis then r_i=†x_i§

I=⁄_im_ir_i2=⁄_im_ix_i2=4µ32+6µ42+8µ0=132 kmÿm2 The kinetic energy is

K= 1

2Iw 2₌1

2132µ12

2₌_{9500 J.}

(b) If the axis is the z-axis then r_i= x_i2+y_i2.

I=⁄_im_ir_i2=⁄_im_iIx_i2+y_i2M=4µI32+52M+6µI42+0M+8µI0+32M=304 kmÿm2 K= 1 2Iw 2₌ 1 2304µ12 2₌_{21 900 J.}

Problem H.8

Two identical thin rods of length L and mass M are perpendicular and joined at their center. What is the moment of inertia about an axis at the end of one rod that is:

(a) perpendicular to the plane, (b) parallel to the other rod.

axis

HaL

axis

HbL

Solution to H.8

The rod connected to the axis has, in both cases, a moment of I₁=1

3M L 2_. The other rod's moment varies.

(a) Use the parallel axis theorem to find the other moment. The distance from the center of mass to the axis is D=Lê2. I2=Icm+M D2= ₁₂1 M L2+MJL₂N2=1₃M L2.

The total moment is the sum.

I=I₁+I₂= 2

3M L 2_.

(b) In this case all the mass of the second rod is a distance Lê2 rom the axis. It's moment is I₂=MJL 2N 2 =1 4M L 2_._ï_I₌_I 1+I2=J1₃+1₄NM L2= ₁₂7 M L2

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Problem H.9

What are the moments of inertia of the following flat uniform bodies of mass M about the axis mentioned? (a) an LµL square about a diagonal

(b) a circle of radius R about a diameter (c) a circle of radius R about a tangent

axis HaL axis HbL axis HcL Solution to H.9

(a) The moment of a flat aµb rectangle of mass M about a perpendicular axis through the center is I= 1

12MIa

2₊_b2_{M. This} follows from the perpendicular axis theorem. (For a flat body in xy-plane, Iz=Ix+Iy) Here take the x and y axes to axes

through the center that are parallel to the sides. For Ix and Iy the moment is equivalent to a uniform rod about a

perpendicu-lar axis through the center. It follows that for a square Iz= 1

6M L

2_{. Now take the}_x_and_y_{axes to be the diagonals. By symmetry}_I

x=Iy=I and by the

perpendicular axis theorem

Iz=Ix+Iy ï I= 1

2Iz= 1 12M L

2

(b) Iz=1₂M R2. Now take the x and y axes to be the perpendicular diameters.. By symmetry Ix=Iy=I and by the

perpen-dicular axis theorem

Iz=Ix+Iy ï I= 1

2Iz= 1 4M R

2

(c) Use the parallel axis theorem to find this moment in terms of the previous. . The distance from the center of mass to the axis is D=R. The axis parallel to the tangent through the center is the diameter, so Icm is the answer to part (b).

I=Icm+M D2= 1 4M R 2₊_{M R}2₌5 4M L 2_.

Problem H.10

(a) A uniform solid cylinder of mass m and radius R rotates about an axis at its rim parallel to the central axis. What is its moment of inertia.

(b) What is the moment of inertia of a uniform solid sphere of mass m and radius R about an axis tangent to its surface. (c) What is the moment of inertia of a uniform thin-shelled hollow sphere of mass m and radius R about an axis tangent to its surface.

Solution to H.10

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(a) I=Icm+M D2=1 2m R 2₊_{m R}2₌ 3 2m R 2_. (b) I=Icm+M D2=2 5m R 2₊_{m R}2₌ 7 5m R 2_. (c) I=Icm+M D2=2 3m R 2₊_{m R}2₌ 5 3m R 2_.