<Chap.
10 Rotational motion: Suggested Problems and Solutions>
10.57, 10.65, 10.69, 10.70, 10.73, 10.78
10.57
A 3.00-kg block rests on a slope and is attached by a string of
negligible mass to a solid drum of mass 0.90 kg and radius 4.50
cm. When released, the block accelerates down the slope at 1.90
𝑀𝑀𝑅𝑅 2
𝑚𝑚/𝑠𝑠 2 . Find the coefficient of friction between block and slope. (𝑓𝑓𝑘𝑘 = 𝜇𝜇𝜇𝜇, 𝐼𝐼 = , 𝑔𝑔 =
2
9.80𝑚𝑚/𝑠𝑠 , √3 = 1.73)
2
Solution)
10pt.
Equation of motion
𝑚𝑚𝑚𝑚 sin 𝜃𝜃 − 𝑓𝑓𝑘𝑘 − 𝑇𝑇 = 𝑚𝑚𝑚𝑚
�
−𝑚𝑚𝑚𝑚 cos 𝜃𝜃 + 𝑁𝑁 = 0
∴ 𝑚𝑚𝑚𝑚 sin 𝜃𝜃 − 𝜇𝜇𝜇𝜇𝜇𝜇 cos 𝜃𝜃 − 𝑇𝑇 = 𝑚𝑚𝑚𝑚
10pt.
wheel
𝜏𝜏𝑛𝑛𝑛𝑛𝑛𝑛 = 𝐼𝐼𝐼𝐼 = 𝑇𝑇𝑇𝑇
𝑀𝑀𝑅𝑅 2 𝑎𝑎
𝐼𝐼𝐼𝐼 𝑀𝑀𝑀𝑀
𝑇𝑇 = = 2 𝑅𝑅 =
𝑅𝑅 𝑅𝑅 2
10pt.
coefficient of friction
𝑀𝑀𝑀𝑀
𝑚𝑚𝑚𝑚 sin 𝜃𝜃 − 𝑚𝑚𝑚𝑚 −
𝜇𝜇 = 2
𝑚𝑚𝑚𝑚 cos 𝜃𝜃
(3.0𝑘𝑘𝑘𝑘)(9.8𝑚𝑚/𝑠𝑠2 ) sin 30° − (3.0𝑘𝑘𝑘𝑘 + 0.45𝑘𝑘𝑘𝑘)(1.9 𝑚𝑚/𝑠𝑠2 )
𝜇𝜇 = = 0.29
(3.0𝑘𝑘𝑘𝑘)(9.8𝑚𝑚/𝑠𝑠2 ) cos 30°
10.65
A disk of radius R has an initial mass M. Then a hole of radius R/4 is drilled, with its edge at the
disk center (Fig. 10.29). Find the new rotational inertia about the central axis. (Hint: Find the
rotational inertia of the missing piece, and substract it from that of the whole disk. You’ll find the
parallel-axis theorem helpful.)
R
R/4
Fig. 10.29
Solution)
10pt.
𝐼𝐼 = � 𝑟𝑟 2 𝑑𝑑𝑑𝑑
20pt.
𝑀𝑀
𝑑𝑑𝑑𝑑 = 𝜎𝜎𝜎𝜎𝜎𝜎 = 2𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑
𝜋𝜋𝑅𝑅2
𝑅𝑅
𝑀𝑀
𝐼𝐼 = � 𝑟𝑟 2 𝑑𝑑𝑑𝑑 = � 𝑟𝑟 2 ∙ ∙ 2𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑
0 𝜋𝜋𝑅𝑅2
2𝑀𝑀 𝑅𝑅 3 1
= 2
� 𝑟𝑟 𝑑𝑑𝑑𝑑 = 𝑀𝑀𝑅𝑅2
𝑅𝑅 0 2
20pt.
𝐵𝐵𝐵𝐵 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑀𝑀 1 2 3 𝑀𝑀𝑅𝑅2 3𝑀𝑀𝑅𝑅2
𝐼𝐼ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = 𝐼𝐼ℎ𝑜𝑜𝑜𝑜𝑜𝑜(𝐶𝐶𝐶𝐶) + ∙ � 𝑅𝑅� = ∙ =
16 4 2 256 512
10pt.
253
𝐼𝐼𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐼𝐼 − 𝐼𝐼ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑀𝑀𝑅𝑅2
512
10.69
A disk of radius 𝑅𝑅 and thickness 𝑤𝑤 has a mass density that increases from the center outward,
given by 𝜌𝜌 = 𝜌𝜌0 𝑟𝑟/𝑅𝑅, where r is the distance from the disk axis. Calculate (a) the disk’s total mass 𝑀𝑀
and (b) its rotational inertia about its axis in terms of 𝑀𝑀 and 𝑅𝑅.
Solution)
(a)
10pt
𝑀𝑀 = � 𝑑𝑑𝑑𝑑 = � 𝜌𝜌𝜌𝜌𝜌𝜌
10pt.
𝑑𝑑𝑑𝑑 = (2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋)𝑤𝑤
𝜌𝜌0 𝑟𝑟
𝜌𝜌𝜌𝜌𝜌𝜌 = � � (2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋)𝑤𝑤
𝑅𝑅
10pt.
𝑅𝑅
𝜌𝜌0 𝑟𝑟
𝑀𝑀 = � � � (2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋)𝑤𝑤
0 𝑅𝑅
𝜌𝜌0 𝑅𝑅
𝜌𝜌0 𝑅𝑅3
𝑀𝑀 = ∙ 2𝜋𝜋𝜋𝜋 � 𝑟𝑟 2 𝑑𝑑𝑑𝑑 = ∙ 2𝜋𝜋𝜋𝜋 ∙
𝑅𝑅 0 𝑅𝑅 3
2𝜋𝜋𝜋𝜋𝜌𝜌0 𝑅𝑅2
∴ 𝑀𝑀 =
3
(b)
10pt.
𝐼𝐼 = � 𝑟𝑟 2 𝑑𝑑𝑑𝑑
10pt.
𝜌𝜌0 𝑟𝑟
𝑏𝑏𝑏𝑏 (𝑎𝑎), 𝑑𝑑𝑑𝑑 = 𝜌𝜌𝜌𝜌𝜌𝜌 = � � (2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋)𝑤𝑤
𝑅𝑅
𝜌𝜌0 𝑅𝑅
𝜌𝜌0 𝑅𝑅5 2𝜋𝜋𝜋𝜋𝜌𝜌0 𝑅𝑅4
𝐼𝐼 = ∙ 2𝜋𝜋𝜋𝜋 � 𝑟𝑟 4 𝑑𝑑𝑑𝑑 = ∙ 2𝜋𝜋𝜋𝜋 ∙ =
𝑅𝑅 0 𝑅𝑅 5 5
10pt.
2𝜋𝜋𝜋𝜋𝜌𝜌0 𝑅𝑅2
𝑀𝑀 =
3
2𝜋𝜋𝜋𝜋𝜌𝜌0 𝑅𝑅4 3
𝐼𝐼 = = 𝑀𝑀𝑅𝑅2
5 5
10.70
The disk in Fig. 10.29 is rotating freely about frictionless horizontal axle. Since the disk is
unbalanced, its angular speed varies as it rotates. If the maximum angular speed is 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 , find an
expression for the minimum speed. (Hint : How does potential energy changes as the wheel
253
rotates?) (rotational inertia is 𝑀𝑀𝑅𝑅2 )
512
R
R/4
Fig 10.29
Solution)
20pt.
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
�⃗ − 1 𝑀𝑀 ∙ 1 𝑅𝑅𝚤𝚤̂
∫ 𝑟𝑟⃗𝑑𝑑𝑑𝑑 𝑀𝑀 ∙ 0
= 16 4 = − 𝑅𝑅 𝚤𝚤̂
∫ 𝑑𝑑𝑑𝑑 𝑀𝑀 60
𝑀𝑀 −
16
20pt.
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚
1 2 1 2
𝐼𝐼𝜔𝜔 + 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚
2 𝑚𝑚𝑚𝑚𝑚𝑚 2
20pt.
1 2 1 2
𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 − 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚
2 2
15 𝑅𝑅 1 2 1 2
𝑀𝑀𝑀𝑀 = 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 − 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚
16 30 2 2
1 2 1 2 𝑀𝑀𝑀𝑀𝑀𝑀
𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐼𝐼𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 −
2 2 32
512 𝑀𝑀𝑀𝑀𝑀𝑀
𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 = �𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚
2 −
253𝑀𝑀𝑅𝑅2 16
32𝑔𝑔
∴ 𝜔𝜔𝑚𝑚𝑖𝑖𝑖𝑖 = �𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚
2 −
253𝑅𝑅
10.73
Calculate the rotational inertia of a solid, uniform right circular cone of mass M, height h, and
base radius R about its axis.
Solution)
10pt.
𝐴𝐴ℎ 𝜋𝜋𝑅𝑅3 ℎ 𝑅𝑅𝑅𝑅
Volume of cone 𝑉𝑉 = = , 𝑑𝑑𝑉𝑉 = 𝜋𝜋𝑅𝑅2 𝑑𝑑𝑑𝑑, 𝑟𝑟 =
3 3 ℎ
𝑀𝑀
Cone has uniform mass density
𝑉𝑉
𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀
𝑑𝑑𝑚𝑚 = = × 𝜋𝜋𝑟𝑟2 𝑑𝑑𝑑𝑑
𝑉𝑉 𝜋𝜋𝑅𝑅2 ℎ
3
10pt.
Inertia of cone
𝑅𝑅 𝑅𝑅 2𝑀𝑀𝑀𝑀 2𝑀𝑀 𝑅𝑅 1
𝐼𝐼 = ∫0 𝑟𝑟2 𝑑𝑑𝑑𝑑 = ∫0 𝑟𝑟2 ( 𝑅𝑅2 )𝑑𝑑𝑑𝑑 = 𝑅𝑅2
∫0 𝑟𝑟3 𝑑𝑑𝑑𝑑 = 2 𝑀𝑀𝑅𝑅2
20pt.
Rotational inertia of solid
𝑟𝑟2 𝑑𝑑𝑑𝑑
𝑑𝑑𝐼𝐼 =
2
ℎ 1 ℎ 3𝑀𝑀 ℎ 𝑅𝑅𝑅𝑅 2 3𝑀𝑀𝑅𝑅2 ℎ
So 𝐼𝐼 = ∫0 𝑑𝑑𝑑𝑑 = ∫0 𝑟𝑟2 𝑑𝑑𝑑𝑑 = 3 ∫0 𝑥𝑥2 ( ℎ ) 𝑑𝑑𝑑𝑑 = 5 ∫0 𝑥𝑥4 𝑑𝑑𝑑𝑑
2 2ℎ 2ℎ
3
∴ 𝐼𝐼 = 𝑀𝑀𝑅𝑅2
10
10.78
Figure shows an object of mass M with one axis through its
center of mass and a parallel axis through an arbitrary point
A. Both axes are perpendicular to the page. The figure
shows an arbitrary mass element dm and vectors
connecting the center of mass, the point A, and dm. (a) Use
�⃗ ∙
the law of cosines to show that 𝑟𝑟 2 = 𝑟𝑟 2 𝑐𝑐𝑐𝑐 + ℎ2 − 2 ℎ
2
𝑟𝑟������⃗
𝑐𝑐𝑐𝑐 . (b) Use this result in 𝐼𝐼 = ∫ 𝑟𝑟 𝑑𝑑𝑑𝑑 to calculate the object’s rotational inertia about the axis
through A. Each term in your expression for 𝑟𝑟 2 leads to a separate integral. Identify one as the
2
rotational inertia about the CM, another as the quantity 𝑀𝑀ℎ , and argue that the third is zero.
Solution)
(a)
10pt.
𝐶𝐶 2 = 𝐴𝐴2 + 𝐵𝐵2 − 2𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 (𝛾𝛾 is the angle between line segments A and B.)
𝑟𝑟 2 = 𝑟𝑟 2 𝑐𝑐𝑐𝑐 + ℎ2 − 2ℎ𝑟𝑟𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
10pt.
�⃗ ∙ 𝑟𝑟������⃗
ℎ𝑟𝑟𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = ℎ 𝑐𝑐𝑐𝑐
�⃗ ∙ 𝑟𝑟������⃗
∴ 𝑟𝑟 2 = 𝑟𝑟 2 𝑐𝑐𝑐𝑐 + ℎ2 − 2 ℎ 𝑐𝑐𝑐𝑐
(b)
10pt.
�⃗ ∙ 𝑟𝑟������⃗
𝐼𝐼 = � 𝑟𝑟2 𝑑𝑑𝑑𝑑 = � 𝑟𝑟 2 𝑐𝑐𝑐𝑐 + ℎ2 − 2 ℎ 𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑
10pt.
� 𝑟𝑟2 𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑 = 𝐼𝐼𝑐𝑐𝑐𝑐
2 2
� ℎ 𝑑𝑑𝑑𝑑 = 𝑀𝑀ℎ
10pt.
∫ 2 �ℎ⃗ ∙ �������⃗𝑑𝑑𝑑𝑑
𝑟𝑟𝑐𝑐𝑐𝑐 = 2 �ℎ⃗ ∙ ∫ �������⃗𝑑𝑑𝑑𝑑
𝑟𝑟𝑐𝑐𝑚𝑚 = 0 (distance to center of mass(CM) from CM is 0)
10pt.
∴ 𝐼𝐼 = 𝐼𝐼𝑐𝑐𝑐𝑐 + 𝑀𝑀ℎ2
