有關普通物理學的問題...

14.During a panic stop,a car decelerates at 7.00 m/s2.(a)What is the angular acceleration of its 0.280 m radius tires,assuming they do not slip on the pavement?(b)How many revolutions do the tires make before coming to rest,given their initial angular velocity is 95.0 rad/s?(c)How long does it take?(d)What distance does the car travel in this time?(e)What was its initial velocity?(f)Do the values obtained seem reasonable,considering that this is a panic stop?

15.The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 × 103 N with an effective perpendicular lever arm of 3.00 cm,producing an angular acceleration of the forcearm of 120 rad/s2. What is the moment of inertia of the boxer's forearm?

16.Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0 kg grindstone(a solid disk).(a)What torque is exerted?(b)What is the angular acceleration assuming negligible opposing friction?(c)What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

17.What is the moment of inertia of an object that rolls without slipping down a 2.00 m high incline starting from rest,and has a final velocity of 6.00 m/s?Express the moment of inertia as a multiple of MR2,where M is the mass of the object and R is its radius.

18.(a)Calculate the angular momentum of the earth in its orbit around the sun.(b)Compare this with the angular momentum of the earth on its axis.

19.Twin skaters approach one another as shown in Figure 9.30 and lock hands.(a)Calculate their final angular velocity,given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg,and their centers of mass are 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius.(b)Compare the initial and final kinetic energy.

2 Answers

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  • 1 decade ago
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    14.題目與4.一模一樣

    A) a= rα → 7=0.28*α → α= 25 rad/s²

    B) ωt²=ω0²+2αθ→ 0=95²-2*25*θ→θ=180.5 rad=28.73 rev

    C) ωt=ω0+αt → 0=95-25*t → t=3.8 sec

    D) S = rθ=0.28*180.5= 50.54 m

    E) V= rω=0.28*95=26.6 m/sec

    F) 不合理,緊急煞車滑行了50多米,耗時近4秒.

    15.τ= Iα;τ= F*r → 2*10³*0.03= I*120 → I=0.5 Kg-m²

    16. a) τ= F*r= 180*0.28=50.4 NT-m

    b) I=mR²/2=75*0.28²/2=2.94 Kg-m²

    τ= Iα→ 50.4=2.94*α→ α=17.14 rad/s²

    c) τ=50.4-20*0.015=50.1 NT-m

    τ= Iα→ 50.1=2.94*α→ α=17.04 rad/s²

    17.無滑動→無摩擦力作功→能量不滅

    V=R*ω→ 6=Rω→ω=6/R

    Mgh=Iω²/2+MV²/2→ M*9.8*2=I*(6/R)²/2 +M*6²/2→ I= 0.089MR²

    18. a) 地球質量5.98*10^24 kg;地球半徑=6.371*10^6m; 地球公轉軌道半徑=1.5*10^11 m

    I≒M*R²=(5.98*10^24)*( 1.5*10^11)²=1.346*10^47 kg-m²

    ω=2π/(365*86400)=1.99*10^-7 rad/s

    角動量=Iω=(1.346*10^47)*( 1.99*10^-7)=2.682*10^40 kg-m²/s

    b) I=MR²/2=(5.98*10^24)*( 6.371*10^6)²/2=2.43*10^38 kg-m²

    ω=2π/86400=7.27*10^-5 rad/s

    角動量=Iω=(2.43*10^38)* (7.27*10^-5)=1.77*10^34 kg-m²/s…..與a)相差近一百萬倍

    19.沒看到圖,我用猜的:設兩人相互接近速度反向,且平行運動相距1.6m

    a) I=ΣM*R²=2*70*0.8²=89.6 kg-m²

    2*70*2.5*0.8= Iω=89.6*ω → ω=3.125 rad/s

    b) E=Σmv²/2=2*70*2.5²=875 Joule

    E=Iω²/2=89.6*3.125²/2=437.5 Joule….為先前的一半

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    7 years ago

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