π ω = π θ π θ ω π θ π θ ω - 物理學系- 東海大學

Fundamentals of Physics 

Halliday & Resnic 

【CH10】Rotational Motion I 

Homework of Chapter 10: 

3, 5, 7, 9, 13, 17, 21, 25, 27, 31, 33, 35, 37, 39, 43, 47, 49, 51, 55, 59, 63, 66 

東海大學物理系 

【Problem 10-3】 

A diver makes 2.5 revolutions on the way from a 10m high platform to the water. Assuming zero 

initial vertical velocity, find the average angular velocity during the dive. 

一個人從 10m 高的跳水台跳到水裡,過程中轉了 2.5 圈。假設初始垂直速度為零,問跳水 

過程中平均角速度為何? 

1 

Δ y = v t+ gt 

2 

: 0 y 

= 0 

⇒ 

2 

2Δy2(10) t = = = 1.4s 

g 9.8 

(2.5 rev)(2 π rad / rev) 

ω avg = = 11 rad / s 

1.4s 


When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. 

If the distance to the floor is 76 cm and for rotation less than 1 rev, what are the (a) smallest and (b) 

largest angular speeds that cause the toast to hit and then topple to be butter-side down? 

一片烤奶油麵包,意外的被推向櫃臺邊緣,當它掉下來時,同時旋轉。如果距離地面 76cm, 

轉速小於 1rev(意思是土司旋轉不會超過 1 圈),問撞到奶油端時(土司另一端撞到地面), 

何為(a)最小和(b)的最大的角速度? 

1 

: h= gt 

2 

2 

2h2(0.76 m) 

Δ t = = = 0.394 s. 

2 

g 9.8 m/s 

(a) 土司旋轉然後掉到地面時的最小角度為 

π 

Δ θmin 

= 0.25 rev = rad 

2 

 



(掉到地面土司旋轉不超過一圈,且 A 端掉下,旋轉後 B 端撞到地面) 

π 

rad 

Δθmin 

ω 

2 

min = = = 4.0 rad/s 

Δt 

0.394 s 

3π 

Δ θ = 0.75 rev = rad 

2 

(b) max 

(掉到地面土司旋轉不超過一圈,且 A 端掉下,旋轉後 D 端撞到地面) 

3π rad 

Δθmax 


2 

max = = = 12.0 rad/s 

Δt 

0.394 s 



The wheel in Fig. 10-30 has eight equally spaced spokes and a radius of 30 cm. It is mounted on a 

fixed axle and is spinning at 2.5 rev/s. You want to shoot a 20cm long arrow parallel to this axle and 

through the wheel without hitting any of the spokes. Assume that the arrow and the spokes are very 

thin. (a) What minimum speed must the arrow have? (b) Does it matter where between the axle and 

rim of the wheel you aim? If so, what is the best location? 

(圖 10-30) 

圖 10-30,一車輪有 8 個等距離的輪輻,其半徑 30cm。它是安裝在固定的軸上旋轉,2.5 轉 

/秒。你想射出一個 20cm 長的箭,使其通過這種車輪而沒有擊中任何的輪輻。假設箭頭和 

輪輻都非常薄。(a)箭的最低速度為多少?(b)你的目標是否是在哪一處車軸間的輪輻? 

如果有的話,何為最好的位置? 

1 

rev 

:(a) 為了避免箭碰到車輪的輪輻,箭經過車輪的時間必須不超過 Δ t = 8 = 0.05s 

2.5 rev / s 

20cm 

所以,箭的最小速度為 vmin = = 400 cm/ s= 4 m/ s 

0.05s



(b) No—there is no dependence on radial position in the above computation. 



A disk, initially rotating at 120 rad / s , is slowed down with a constant angular acceleration of 

2 

magnitude 4 rad / s . (a) How much time does the disk take to stop? (b) Through what angle does 

the disk rotate during that time? 

2 

一圓盤,一開始轉動速率120 rad / s ,緩慢地以角加速度 4 rad / s 減少轉速。(a)圓盤停止 

時花了多少時間?(b)在那段時間,圓盤轉了多少角度? 

:(a) ω = ω0+ αt 

ω −ω0 0−120 ⇒ t = = = 30s 

α ( −4) 

1 1 

3 

(b) θ = ( ω+ ω0) 

t = (120 + 0)(30) = 1.8× 10 rad 

2 2 


2 

A wheel has a constant angular acceleration of 3 rad / s . During a certain 4.0 s interval, it turns 

through an angle of 120 rad. Assuming that the wheel started from rest, how long has it been in 

motion at the start of this 4.0 s interval? 

2 

一輪子具有固定角加速度 3 rad / s 。在某一個 4秒期間,它轉了 120rad。假設輪子一開始 

是靜止狀態,問在這個 4秒期間之前,它運動了多久? 

1 

: θ = ω0t+ αt 

2 

2 

1 2 1 2 

θ − αt120 

− (3)(4 ) 

ω 2 2 

0 = = = 24 rad / s …在 4 秒期間,一開始時的轉速。 

t 

4 

ω = ω0+ αt 

ω −ω0 24 − 0 

t = = = 8s,輪子一開始靜止,8 

秒後,轉速增加為 24 rad / s 。 

α 3 


A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) 

Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular 

acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions? 

一飛輪,當它從角速度 1.5rad/ s 減速到停止,共旋轉了 40rev。(a)假設角加速度為定值, 

問它到停止時共花了多少時間。(b)角加速度為何?(c)40 個週期的的前 20 週期,需要 

花多少時間? 

 



1.5 

:假設 t = 0 , ω0 

= 1.5 rad / s = rev / s = 0.239 rev / s 


t = t1,轉了 

θ 1 = 20rev 

t = t2,轉了 

θ 2 = 40rev , ω 2 = 0 


1 

(a) θ2 = ( ω0 + ω2) 

t2 

2 

2θ 2 2(40) 

⇒ t2= = = 335s 

ω0 + ω2 

0+ 0.239 

1 2 

(b) θ2 = ω2t2 − αt2 

2 

2( ω2t2 −θ 2) 

2(0 − 40) 

−4 

2 

−2 

2 

⇒ α = = =− 7.13× 10 rev / s =− 4.48× 10 rad / s 

2 2 

t2 

335 

1 2 

(c) θ1 = ω0t1+ αt1 

2 

2 

2 −4 

− ω0 ± ω0 + 2θα 

1 − (0.239) ± (0.239) −2(20)( − 7.13× 10 ) 

t1 

= 

= 

= 98 或 572 

−4 

α 

− 7.13× 10 

∵ t1 < t2 

t = 98s 

∴ 1 


A flywheel with a diameter of 1.2 m is rotating at an angular speed of 200 rev/min. (a) What is the 

angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the 

rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will 

increase the wheel’s angular speed to 1000 rev/min in 60s? (d) How many revolutions does the 

wheel make during that 60s? 

飛輪直徑 1.2 米,旋轉角速度為 200 轉 /分。(a)寫出飛輪角速度以弧度每秒?(b)飛輪 

邊緣上一個點的直線速度?(c)固定角加速度為何(轉/分 2 ),60 秒內將飛輪角速度增加 

到到 1000 轉 /分?(d)在該 60 秒內,轉了多少轉? 

200rev 2π 

rad 

( )( ) 

:(a) ω = min rev = 20.9 rad / s 

60s 

( ) 

min 

1.2 

(b) r = = 0.6m 

2 

v= rω= (0.6)(20.9) = 12.5 m/ s 

(c) t = 60s= 1min , 1000 rev / min 

ω = , ω 0 = 

200 rev / min




2 

α = = = 800 rev / min 

t 1 

1 1 

θ = ( ω + ω) 

t = (200 + 1000)(1) = 600rev 

2 2 

(d) 0 



A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting 

from rest, gaining angular speed at the constant rate α 1 through the first 400 rad and then losing 

angular speed at the constant rate − α1 

until it is again at rest. The magnitude of the centripetal 

2 

acceleration of any portion of the disk is not to exceed 400 m/ s . (a) What is the least time 

required for the rotation? (b) What is the corresponding value of α 1 ? 

一張盤子,半徑 0.25 m,從靜止開始被轉動像旋轉木馬一樣的轉了 800 rad,以角加速度 α 1 轉 

了前 400rad。然後以角加速度為 − α1 

減速到停止。盤子任何部分的向心加速度不得超過 

2 

400 m/ s 。(a)轉動需要多少時間?(b) α 1 為何? 

: r = 0.25m 

2 

v 2 

ar= = rω 

r 

ar 

400 

⇒ ω = = = 40 rad / s 

r 0.25 

1 

(a) θ − θ0 = ( ω0 + ω) 

t 

2 

2( θ −θ 0) 

2(400 − 0) 

t = = = 20s……這是一半的時間 

ω0+ ω 0+ 40 

所以花的總時間為 40 秒。 

(b) ω = ω0+ αt 


α = = = 2 rad / s 

t 20 

2 


An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of 

light passes through one of the slots at the outside edge of the wheel, as in Fig. 10-31, travels to a 

distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One 

such slotted wheel has a radius of 5 cm and 500 slots around its edge. Measurements taken when 

5 

the mirror is L= 500m 

from the wheel indicate a speed of light of 3× 10 km / s . (a) What is the 

(constant) angular speed of the wheel? (b) What is the linear speed of a point on the edge of the 

wheel? 

早期測量光速的方法是利用旋轉的槽輪。一道光束穿過車輪邊緣的凹槽,如圖 10-31,射到 

 




一段距離的鏡子,並返回到車輪正好穿過車輪的下一個凹槽。像這樣的一個有凹槽的輪子 

5 

半徑為 5cm 左右且外圍有 500 個凹槽。鏡子距離輪子 L= 500m,量到的光速為 

3× 10 km / s(a) 

車輪的角速度(常數)為何?(b)車輪邊緣的一個點,它的線速度為何? 


−2 

:(a) θ = = 1.26× 10 rad 

500 

2L 2(500 m) 

−6 

t = = = 3.33× 10 s 

8 

c 3× 10 m/ s 

−2 

θ 1.26× 10 rad 

ω = = 

−6 

t 3.33× 10 s 

= × 

3 

3.8 10 / 

rad s 

3 

(b) v= ω r = ( 3.8× 10 rad/s) ( 0.050 m) 

2 

= 1.9× 10 m/s. 

(圖 10-31) 


A record turntable is rotating at 3313 rev/min. A watermelon seed is on the turntable 6.0 cm from 

the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) 

What is the minimum value of the coefficient of static friction between the seed and the turntable if 

the seed is not to slip? (c) Suppose that the turntable achieves its angular speed by starting from rest 

and undergoing a constant angular acceleration for 0.25 s. Calculate the minimum coefficient of 

static friction required for the seed not to slip during the acceleration period. 

一個記錄轉盤旋轉 3313 轉/分鐘,一顆西瓜種子在轉盤上,離轉軸 6cm。(a)如果種子不滑 

動,求出種子的加速度(b)如果種子不滑動,求出種子與轉盤的最小靜摩擦係數(c)假設 

轉盤從靜止,花了 0.25s 等速加速到一個固定角速度,加速過程中,種子不滑動,求出最小 

靜摩擦係數。 

:(a) The angular speed in rad/s is



ω = F H G 

33 1 

rev / min 

3 

I K J F H G 

I 


rad / rev 

= 349 . rad / s. 

60 s/minKJ 

Consequently, the radial (centripetal) acceleration is (using Eq. 10-23) 

b g 

− 

a = ω r = × = 

2 2 2 

349 . rad / s ( 60 . 10 m) 0.73 m / s . 

2 


(b) Using Ch. 6 methods, we have ma = fs ≤ fs,max = μs mg, which is used to obtain the 

(minimum allowable) coefficient of friction: 

a 073 . 

μ s,min 

= = = 0075 . . 

g 98 . 

(c) The radial acceleration of the object is ar = ω 2 r, while the tangential acceleration is at = 

αr. Thus, 

2 2 2 2 2 4 2 

| a| = a + a = ( ω r) + ( αr) = r ω + α . 

r t 

If the object is not to slip at any time, we require 

4 2 

f = μ mg = ma = mr ω + α 

s,max s 

max max . 

Thus, since α = ω/t (from Eq. 10-12), we find 

μ 

s,min 

4 2 4 2 4 2 

r ωmax + α r ωmax + ( ωmax 

/ t) 

(0.060) 3.49 + (3.4/0.25) 

= = = = 0.11. 

g g 

9.8 


Calculate the rotational inertia of a wheel that has a kinetic energy of 24 400 J when rotating at 602 

rev/min. 

車輪具有動能 24,400J,轉速 602 轉/分,計算車輪的轉動慣量。 

1 2 

: K = Iω 

2 

2K 2(24,400 J ) 

I = = = 12.3kg 

⋅ m 

2 

ω (602 rev / min)(2 π rad / rev) 

60 s / min 


Calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to 

the stick and located at the 20 cm mark. (Treat the stick as a thin rod.) 

計算一米棒子的轉動慣量,其質量 0.56kg,有一軸垂直於棒子,固定在 20cm 位置。(把棒 

子視如一隻細竿子。) 

2 

 



:依據平行軸定理: I = I + Mh 

h= 0.5 − 0.2 = 0.3m 

CM 

2 

1 1 

I = ICM + Mh = ML + Mh 

= (0.56)( 1) + (0.56)(0.3) 

= 9.7× 10 kg 

⋅ m 

12 12 

2 2 2 

2 

2 

−2 

2 



Two uniform solid cylinders, each rotating about its central (longitudinal) axis at 235 rad/s, have the 

same mass of 1.25 kg but differ in radius. What is the rotational kinetic energy of (a) the smaller 

cylinder, of radius 0.25 m, and (b) the larger cylinder, of radius 0.75 m? 

二個一樣的堅實圓筒,互相對縱向中間軸轉動,轉速 235 rad/s,有相同質量 1.25 kg,但是 

不同的半徑。(a)小圓筒半徑 0.25 m 和(b)大圓筒半徑半徑 0.75 m,問兩者的轉動動能? 

1 2 

: I = MR 

2 

1 2 1 2 2 

K = Iω= MR ω 

2 4 

1 2 2 1 2 2 3 

(a) K = MR ω = (1.25)(0.25) (235) = 1.1× 10 J 

4 2 

1 2 2 1 2 2 3 

(b) K = MR ω = (1.25)(0.75) (235) = 9.7 × 10 J 

4 2 


In Fig. 10-36, two particles, each with mass m= 0.85kg 

, are fastened to each other, and to a 

rotation axis at O, by two thin rods, each with length d = 5.6cm 

and mass M = 1.2kg 

. The 

combination rotates around the rotation axis with angular speed ω = 0.3 rad / s . Measured about O, 

what are the combination’s (a) rotation inertia and (b) kinetic energy? 

圖 10-36,兩個粒子,質量都是 m= 0.85kg 

,被綁在一起。由兩個細棒對 O 點旋轉,細棒每 

個長度 d = 5.6cm和質量 

M = 1.2kg 

。組合的旋轉軸角速度 ω = 0.3 rad / s 。問組合的:(a)轉 

動慣量和(b)動能?



(圖 10-36) 

: d = 5.6cm= 0.056m 

(a) I = I1+ I2 + I3 + I4 

⎛ 1 1 ⎞ 2 ⎛ 1 2 3 2⎞ 

= ⎜ Md + M( d) ⎟+ md + ⎜ Md + M( d) 

⎟+ 

m( 2 d) 

⎝12 

2 ⎠ ⎝12 2 ⎠ 

(b) 

2 2 2 

8 2 2 8 2 2 

= Md + 5md = (1.2)(0.056) + 5(0.85)(0.056) 

3 

3 

1 2 

−3 

K = Iω 

= 1.04× 10 J 

2 

1 2 

= (0.023)(0.3) 

2 

2 

= 0.023kg ⋅ m 



Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the 

flywheel up to its top speed of 200rad/s. One such flywheel is a solid, uniform cylinder with a mass 

of 500 kg and a radius of 1.0 m. (a) What is the kinetic energy of the flywheel after charging? (b) If 

the truck uses an average power of 8.0 kW, for how many minutes can it operate between 

chargings? 

能量儲存在一個旋轉的飛輪使得卡車可以運行。電動馬達使飛輪速度升到最大 200rad / s。 

這樣的飛輪是堅固的,制式汽缸質量 500kg,半徑 1m。(a)輪子轉動後的動能為何?(b) 

假如卡車平均功耗為 8 千瓦,要花多少分鐘才能運轉? 

:(a) Using Table 10-2(c) and Eq. 10-34, the rotational kinetic energy is 

1 2 1⎛1 2⎞ 2 1 2 2 7 

K = Iω = ⎜ MR ⎟ω 

= (500kg)(200 π rad/s) (1.0m) = 4.9× 10 J. 

2 2⎝2 ⎠ 4 

(b) We solve P = K/t (where P is the average power) for the operating time t. 

t K 

7 

49 . × 10 J 

3 

= = 

= 62 . × 10 s 

3 

P 8.0 × 10 W 

which we rewrite as t ≈ 1.0 ×10 2 min. 


The body in Fig. 10-39 is pivoted at O, and two forces act on it as shown. If 1 1.3 r m = , r2= 2.15m, 

0 

0 

F = 4.2N 

, F2 = 4.9N 

, θ = 75 , and θ = 60 , what is the net torque about the pivot? 

1 

1 

2 

一物體如圖 10-39,以 O 點為軸(固定在 O 點),有兩個力量作用在上面,如果, r1= 1.3m, 

0 

0 

r = 2.15m, 

F1 = 4.2N 

, F2 = 4.9N 

, θ = 75 和 θ = 60 ,對 O 點的淨力矩為何? 

2 

1 

2 

(圖 10-39) 

 



: τ = rF 1 1sinθ1− rF 2 2sinθ2 0 0 

= (1.3)(4.2)sin 75 − (2.15)(4.9)sin 60 

= −3.85N⋅ m 



During the launch from a board, a diver’s angular speed about her center of mass changes from zero 

2 

to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is 12kg ⋅ m . During the 

launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external 

torque on her from the board? 

從地板跳水的過程中,一位跳水者質量中心的角速度,在 220ms 內,從零變到 6.20 弧度/ s。 

2 

質量中心的轉動慣量為 12kg ⋅ m 。在跳水過程中,(a)她的平均角加速度為何?(b)從地 

板,平均力矩為何? 

:(a) ω = ω0+ αt 

ω −ω0 6.2 − 0 

⇒ α = = = 28.2 rad / s 

−3 

t 220× 10 

2 

(b) τ = Iα = (12)(28.2) = 3.38× 10 N⋅ m 


Figure 10-41 shows a uniform disk that can rotate around its center like a merry-go-round. The disk 

has a radius of 2 cm and a mass of 20grams and is initially at rest. Starting at time t = 0 , two forces 

are to be applied tangentially to the rim as indicated, so that at time t = 1.25s 

the disk has an 

angular velocity of 250 rad / s counterclockwise. Force F1 has a magnitude of 0.1 N. What is 

magnitude 2 F ? 

圖 10-41 顯示了一個標準盤子,可以繞著它的中心轉動,就像旋轉木馬。盤子半徑為 2cm, 

質量 20g,一開始是靜止的。時間 t = 0 開始轉動,兩個力作用在外圍切線面上,在 t = 1.25s 

時,盤子角速度為逆時針 250 rad / s 。 F1 力大小為 0.1N。問 F2 力大小為何? 

: τ net = Iα 

RF2 − RF1 = Iα 

ω = ω + αt⇒ 

0 

= 0 


α = 

t 

2 

(圖 10-41)



1 2 

I = MR 

2 

Iα 

1 2 ω 1 MRω 

(0.02)(0.02)(250) 

F2 = F1+ = F1+ ( MR )( )( ) = F1 + = 0.1+ = 0.14N 

R 2 t R 2t2(1.25) 東海大學物理系 


In Fig. 10-45, block 1 has mass m1= 460g, 

block 2 has mass m2= 500g, 

and the pulley, which is 

mounted on a horizontal axle with negligible friction, has radius R = 5cm 

. When released from rest, 

block 2 falls 75.0 cm in 5.00 s without the cord slipping on the pulley. (a) What is the magnitude of 

the acceleration of the blocks? What are (b) tension T 2 and (c) tenson T 1? 

(d) What is the 

magnitude of the pully’s angular acceleration? (e) What is its rotational inertia? 

圖 10-45。積木 1 質量 m1= 460g,積木 

2 質量 m2= 500g,被掛在一無摩擦滑輪上,滑輪半 

徑 R = 5cm 

。從靜止被釋放,積木在 5 秒內降落了 75 公分,繩子沒有滑動。(a)積木加速 

T ?(d)滑輪角加速度?(e)滑輪轉動慣量? 

度?(b)繩子張力 T 2 ?(c)繩子張力 1 

(圖 10-45) 

:(a) We use constant acceleration kinematics. If down is taken to be positive and a is the 

acceleration of the heavier block m2, then its coordinate is given by y = at 1 2 

2 , so 

2y2(. 0750m) 

−2 

2 

a = = = 600 . × 10 m/s . 

2 2 

t (. 500s) 

Block 1 has an acceleration of 6.00 × 10 –2 m/s 2 upward. 

(b) Newton’s second law for block 2 is mg 2 − T2 = ma 2 , where m2 is its mass and T2 is the 

tension force on the block. Thus, 

2 2 

2 −2 

2 

( ) 

T = m ( g− a) 

= (0.500 kg) 9.8 m/s − 6.00× 10 m/s = 4.87 N. 

(c) Newton’s second law for block 1 is 1 1 1 , 

mg T ma 

the block. Thus, 

1 1 

− = − where T1 is the tension force on 

2 −2 

2 

( ) 

T = m ( g+ a) 

= (0.460 kg) 9.8 m/s + 6.00× 10 m/s = 4.54 N. 

(d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the 

 



rim of the pulley must be the same as the acceleration of the blocks, so 

−2 

2 

a 600 . × 10 m/s 

2 

α = = 

= 120 . rad / s . 

−2 

R 500 . × 10 m 


(e) The net torque acting on the pulley is τ = ( T2 − T1) R. 

Equating this to Iα we solve for the 

rotational inertia: 

−2 

( T −T 

) R ( − )( × ) 

4.87 N 4.54 N 5.00 10 m 2 1 −2 

2 

I = = = 1.38× 10 kg ⋅ m . 

2 

α 

1.20 rad/s 


A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be 

brought to a stop in 15.0 s. (a) How much work must be done to stop it? (b) What is the required 

average power? 

一個 32 公斤輪子,基本上是一個薄圈,半徑 1.2m,每分鐘 280 轉。它必須在 15 秒內停止 

(a)為了使它停止需作多少功?(b)所需的平均功率為何? 

280rev 2π 

: ω = ( ) = 29.3 rad / s 

min 60 

2 2 2 

(a) 轉動慣量: I = mr = (32)(1.2) = 46.1kg 

⋅ m 

1 2 1 

2 4 

W =Δ K = 0 − Iω=− (46.1)(29.3) = 1.98× 10 J 

2 2 

3 

W 19.8× 10 

3 

(b) P = = = 1.32× 10 W 

Δt 

15 


A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed 

of the other end just before it hits the floor, assuming that the end on the floor does not slip. (Hint: 

Consider the stick to be a thin rod and use the conservation of energy principle.) 

一米長棒子,直立的讓一端固定在地板,使其落下,問另一端撞到地面的速度。假設固定 

的那端不滑動。(提示:考慮棒子是細長的,且利用能量守恆原理。) 

:We use to denote the length of the stick. Since its center of mass is /2 from either end, 

its initial potential energy is 1 

2 mg, where m is its mass. Its initial kinetic energy is zero. Its 

final potential energy is zero, and its final kinetic energy is 1 2 

Iω , where I is its rotational 

inertia about an axis passing through one end of the stick and ω is the angular velocity just 

before it hits the floor. Conservation of energy yields 

2




1 1 2 mg 

mg = Iω 

⇒ ω = . 

2 2 I 

The free end of the stick is a distance from the rotation axis, so its speed as it hits the 

floor is (from Eq. 10-18) 

v 

= ω = 

mg 

I 

3 

. 

Using Table 10-2 and the parallel-axis theorem, the rotational inertial is I = m 1 

2 

v = 3g = 3 9.8 m / s 1.00 m = 5.42 m / s. 

c hb g 

3 

2 , so 


A uniform spherical shell of mass M = 4.5kg 

and radius R = 8.5cm 

can rotate about a vertical 

axis on frictionless bearings (Fig. 10-48). A massless cord passes around the equator of the shell, 

−3 

2 

over a pulley of rotational inertia I = 3× 10 kg ⋅ m and radius r = 5cm, 

and is attached to a small 

object of mass m= 0.6kg 

. There is no friction on the pulley’s axle; the cord does not slip on the 

pulley. What is the speed of the object when it has fallen 82 cm after being released from rest? Use 

energy considerations. 

一均勻球殼,質量 M = 4.5kg 

,半徑 R = 8.5cm 

,繞著無摩擦的軸承上的鉛直軸轉動。如圖 

−3 

2 

10-48,一繩子繫在球的赤道上,經由轉動慣量 I = 3× 10 kg ⋅ m ,半徑 r = 5cm的滑輪,與質 

量 m= 0.6kg 

的小物體連接。滑輪上無摩擦,繩子在滑輪上沒有滑動,當物體從靜止降落了 

82cm 時,物體速度為何? 

(圖 10-48) 

2 2 

:球殼轉動慣量為 

3 MR 

so the kinetic energy (after the object has descended distance h) is 

K = MR I mv 

F HG 1 2 2I2 1 2 1 2 

KJ ωsphere + ω pulley + . 

2 3 

2 2 

Since it started from rest, then this energy must be equal (in the absence of friction) to the 

potential energy mgh with which the system started. We substitute v/r for the pulley’s 

 



angular speed and v/R for that of the sphere and solve for v. 

mgh 2gh 

v = = 

m+ + + I mr + M m 

1 1 I M 

2 2 2 

r 3 

2 

1 ( / ) (2 /3 ) 

2(9.8)(0.82) 

= = 1.4 m/s 

+ × + 

−3 

2 

1 3.0 10 /((0.60)(0.050) ) 2(4.5)/3(0.60) 

東海大學物理系

π ω = π θ π θ ω π θ π θ ω - 物理學系- 東海大學

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