PHYS 101 - General Physics I Final Exam Solution: Duration: 150 Minutes Monday, 5 June 2023 18:00
PHYS 101 - General Physics I Final Exam Solution: Duration: 150 Minutes Monday, 5 June 2023 18:00
PHYS 101 - General Physics I Final Exam Solution: Duration: 150 Minutes Monday, 5 June 2023 18:00
1. A disc of mass 𝑀1 and radius 𝑅 ( 𝐼 = 𝑀1 𝑅2 /2) is wrapped with an ideal string and is
connected to a mass 𝑀2 over a massless and frictionless pulley as shown in the figure.
When the system is released from rest, both the center of the disc and the mass 𝑀2 start
moving down with the same acceleration. Gravitational acceleration is given as g, and
assume that the string tied to the disc stays vertical throughout the motion.
(a) (4 Pts.) Draw the free body diagram for both masses.
(b) (12 Pts.) What is the ratio 𝑀1 /𝑀2 ?
(c) (4 Pts.) What is the angular acceleration of the disc?
𝑀2
ሬ𝑻Ԧ ሬ𝑻Ԧ
Solution: (a, b) Writing Newton’s second law for the
rotational motion of the disc, we have 𝑀1
1 ሬԦ
𝑀1 𝐠
𝑅𝑇 = 𝑀1 𝑅2 𝛼 . ሬሬԦ
𝑀2 𝒈
2
Point of contact of the string with the disc is accelerating up with acceleration 𝑎 and the center of the disk is
accelerating down with the same acceleration 𝑎. Therefore, relative acceleration of the center of the disc with respect
to the string is 2𝑎. Hence
2𝑎
𝛼= → 𝑇 = 𝑀1 𝑎
𝑅
g
𝑀1 g − 𝑇 = 𝑀1 𝑎 → 𝑎=
2
(c)
2𝑎 g
𝛼= → 𝛼= .
𝑅 𝑅
2. A solid, uniform sphere of mass 𝑀 and radius 𝑟0 starts from rest and rolls without slipping down an inclined plane
of height 𝐻, and angle of inclination 𝜃.
Solution:
(a) Total mechanical energy is conserved throughout the motion.
1 1 𝑣c
𝐸i = 𝑀g𝐻 , 𝐸f = 𝑀𝑣c2 + 𝐼𝜔2 , 𝜔=
2 2 𝑟0
1 1 2 𝑣𝑐 2 10
𝐸𝑓 = 𝐸𝑖 → 𝑀𝑣c2 + ( 𝑀𝑟02 ) ( ) = 𝑀g𝐻 → 𝑣𝑐 = √ g𝐻 .
2 2 5 𝑟0 7
(b)
𝑀g sin 𝜃 − 𝐹𝑓𝑟 = 𝑀𝑎 , 𝐹𝑁 − 𝑀g cos 𝜃 = 0
2 2 𝑎 2
𝑟0 𝐹𝑓𝑟 = ( 𝑀𝑟02 ) 𝛼 = ( 𝑀𝑟02 ) → 𝐹𝑓𝑟 = 𝑀𝑎
5 5 𝑟0 5
Tork denklemi r.F=I.𝛼
Rotational 𝛼=a/r0 dönüşümü
2 5
𝑀g sin 𝜃 − 𝑀𝑎 = 𝑀𝑎 → 𝑎 = g sin 𝜃
5 7
mgSin𝜃 yönünde Newton's 2. law
𝑎 5g
𝛼= → 𝛼= sin 𝜃.
𝑟0 7𝑟0
(c)
2 2
𝐹𝑓𝑟 = 𝑀𝑎 = 𝑀g sin 𝜃 , 𝐹𝑁 = 𝑀g cos 𝜃
5 7 Eg. 11.1
𝐹𝑁 = 𝑀g cos 𝜃
𝐹𝑓𝑟 2
𝐹𝑓𝑟 ≤ 𝜇𝑠 𝐹𝑁 → 𝜇𝑠 ≥ → 𝜇𝑠 ≥ tan 𝜃 . 2 2
𝐹𝑁 7 𝐹𝑓𝑟 = 𝑀𝑎 = 𝑀g sin 𝜃 ,
5 7
3. A uniform disk of mass 𝑀 and radius 𝑅 slides without rotating across a frictionless horizontal icy surface at speed
𝑣0 . It makes a glancing collision with an identical disk which is initially at rest such that their rims just touch. Because
their rims are coated with instant acting glue, the pucks stick together following the collision. The moment of inertia of
a uniform disk about its center is 𝐼𝐶 = 𝑀𝑅 2 /2.
(a) (8 Pts.) Find the center of mass speed of the combined
two-disc system after the collision.
𝐯ሬԦ0 𝜔
(b) (12 Pts.) Find the angular speed of rotation around the
center of mass after the collision.
2𝑅
𝑀, 𝑅
Solution: ሬԦ𝐶𝑀
𝒗
Moment of inertia of the two disks about their common center of mass is
1
𝐼 = 2 ( 𝑀𝑅 2 + 𝑀𝑅 2 ) → 𝐼 = 3𝑀𝑅 2
2
Therefore,
𝑣0
𝐿𝑓 = 𝐿𝑖 → 𝜔=
3𝑅
4. A space ship is stranded without any fuel in a circular orbit around a star. The radius of the orbit is 𝑅 and the speed
of the space ship is 𝑣0 .
(a) (6 Pts.) What is the mass of the star?
In order to escape from this obit and get back to our solar system the captain of the ship designs an explosion that will
split the ship into two equal parts both with mass 𝑚/2. After the explosion one half of the ship will remain near the
star, and the other half carrying the passengers will escape from the star system to our solar system (which can be
assumed to be infinitely far away)
(b) (7 Pts.) What should be the minimum speed of the escaping part after the explosion in terms of the given quantities
so that it can get infinitely far away from the star?
(c) (7 Pts.) What is minimum energy of the explosion so that the escaping part can reach the escape speed?
Solution:
(a)
𝑀𝑚 𝑚𝑣02 𝑅𝑣02
𝐹𝐺 = 𝐺 , 𝐹𝐺 = → 𝑀= .
𝑅2 𝑅 𝐺
1 𝑚 2 𝐺𝑀 𝑚
𝐸 = ( ) 𝑣min − ( )
2 2 𝑅 2
1 𝑚 2 𝐺𝑀 𝑚 2 2𝐺𝑀
( ) 𝑣min − ( )=0 → 𝑣min = = 2𝑣02 → 𝑣min = √2𝑣0
2 2 𝑅 2 𝑅
𝑚 𝑚
𝑝𝑖 = 𝑚𝑣0 , 𝑝𝑓 = ( ) 𝑣1 + ( ) 𝑣min → 𝑣1 = (2 − √2)𝑣0 .
2 2
1 1 𝑚 2 1 𝑚 1 1
𝐸𝑖 = 𝑚𝑣02 , 𝐸𝑓 = ( ) (2 − √2) 𝑣02 + ( ) 2𝑣02 = (3 − 2√2)𝑚𝑣02 + 𝑚𝑣02
2 2 2 2 2 2 2
1
Δ𝐸 = (3 − 2√2) 𝑚𝑣02 = (3 − 2√2)𝐸𝑖 .
2
5. (20 Pts.) A bicycle wheel of mass 𝑀 and radius 𝑅 has a moment of inertia 𝐼0 Pivot
around its center of mass, which is at its center. If the wheel is suspended on a
wall from its rim with a frictionless pivot as shown in the figure, find the period
of its small oscillations. Gravitational acceleration is given as g.
Solution:
(1) Use Newton’s second law. ሬԦ
𝐠
𝑑2 𝜃
𝜏 = 𝐼𝛼 → −𝑀g𝑅 sin 𝜃 = (𝐼0 + 𝑀𝑅 2 )
𝑑𝑡 2
𝑇 =?
𝑑2 𝜃 𝑀g𝑅 𝑀g𝑅 2𝜋 𝐼0 + 𝑀𝑅 2
2
+ 2
𝜃=0 →𝜔=√ → 𝑇= = 2𝜋√
𝑑𝑡 𝐼0 + 𝑀𝑅 𝐼0 + 𝑀𝑅 2 𝜔 𝑀g𝑅
𝑀𝑔𝑅
𝜔=√
𝐼𝑃