he surface of a solid ball is glued to one end of a rod and the rod is rotated about an axis through the other end of the rod and perpendicular to it. The ball’s mass is 4.00 kg, its radius is 2.00 m. The rod’s mass is 6.00 kg and its length is 3.00 m. What is the moment of inertia of the system (taking into account the ball’s size)?
he surface of a solid ball is glued to one end of a rod and the rod is rotated about an axis through the other end of the rod and perpendicular to it. The ball’s mass is 4.00 kg, its radius is 2.00 m. The rod’s mass is 6.00 kg and its length is 3.00 m. What is the moment of inertia of the system (taking into account the ball’s size)?
Related questions
Question
Please use provided formulas in sheet and explain step by step don’t copy from chegg please, 17.18,19,20
17. The surface of a solid ball is glued to one end of a rod and the rod is rotated about an axis through the other end of the rod and perpendicular to it. The ball’s mass is 4.00 kg, its radius is 2.00 m. The rod’s mass is 6.00 kg and its length is 3.00 m. What is the moment of inertia of the system (taking into account the ball’s size)?
18. A massless beam of length 4.00 m supports a 200 N ball as in figure. Angle 1 is 300 and angle 2 is 600. What is the tension in the upper supporting wire? answer is 100
19. An Atwood machine (similar to an elevator with a counter-weight) is initially at rest. On one side is a mass of 2.00 kg and on the other side is a mass of 1.00 kg. They are connected by a massless wire that passes over a pulley. The pulley has a mass of 4.00 kg, a radius of 20.0 cm and no friction. The heavier mass then falls for 50.0 cm. What is the linear speed at that point? answer is 1.4
20. A ring is dropped concentrically and with a parallel axis on a rotating table. The ring and the table each have a moment of inertia of 6.00 kg m2. The ring is initially not rotating and the table is initially rotating at 3.00 rad/s. How much energy, in J, is lost to heat as the ring lands on the table? answer is 13.5
Transcribed Image Text:The exam is closed book and closed notes.
v2
Circular motion: ac =
W
Pavr =
At
Fnet = ma;
Weight: Fg = mg, ;
g = 9.8 m/s?;
Kinetic energy: K = -m
1
v?; Potential energy: Ug = mgy
E = Er
E = U+K
v2
= o'r;
Rotational motion:
e =
1 rev = 2n rad;
v = o r;
at = a r;
a, =
0 = 0o + at;
1
e = 0o t +
20a = m? - 0o?
K=
T=rx F;
T= rFsino;
Στ-Ια;
Ipoint mass = mr?
Idisk =
mR 2
mR 2
Rin
1
Ihoop = mR?
2
2
Ishell = =mR?
3
Irod(center) =
mL?
Irod(end) = -mL²
Iball =
mR?
5
12
I= Icom + MD?
work: W =t 0;
dW
P =
dt
2.
W
Payr =
Δt
W =
Rolling:
K =
2+
1
m Vcom
Vcom = Ro
Io
T=f,R
F,max = H,Fn
Incline: Fg= mgsin0 Fg mgcos0
Angular momentum:
Lpoint mass =m rxv
L= mrvsin 0;
L=m (r¿vy - ryVz)k
L= Io
LI = L
I 01 =I 202
m,x1+m2X2
mıyı+m2y2
X com
Y com =
m1 +m2
m1 +m2
1
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 1 images
