Please explain me step by step don’t copy paste from chegg please and use formula as I provided sheet here 10,11,12 10. A mass of 0.50 kg pulls down vertically on a string that is unwinding around a solid clyndrical rod attached to a disk. The rod and the disk are concentric with a horizontal axis. Both are uniform. The disk has a radius of 0.20 m and mass 0.20 kg. The rod has a radius of 0.090 m and mass 2.0 kg and is supported with no friction. What is the moment of inertia, in kg m2, of the combination of rod and disk? answer is 0.012 11.A disk of radius 0.10 m falls vertically with its axis horizontal, unwinding a string whose other end is attached above. The moment of inertia of the disk is 4.0 kg.m2, and its angular acceleration is 2.0 rad/s2. What is the tension in the string?answer is 80 12. A constant torque brings a solid, cylinder grinding wheel from rest to an angular speed of 209 rad/s. The wheel has a mass of 1.8 kg and a radius of 0.13 m. What torque in N m is needed to bring the wheel up to that speed in 2.9 s? answer is 1.1

icon
Related questions
Question
100%
Please explain me step by step don’t copy paste from chegg please and use formula as I provided sheet here 10,11,12 10. A mass of 0.50 kg pulls down vertically on a string that is unwinding around a solid clyndrical rod attached to a disk. The rod and the disk are concentric with a horizontal axis. Both are uniform. The disk has a radius of 0.20 m and mass 0.20 kg. The rod has a radius of 0.090 m and mass 2.0 kg and is supported with no friction. What is the moment of inertia, in kg m2, of the combination of rod and disk? answer is 0.012 11.A disk of radius 0.10 m falls vertically with its axis horizontal, unwinding a string whose other end is attached above. The moment of inertia of the disk is 4.0 kg.m2, and its angular acceleration is 2.0 rad/s2. What is the tension in the string?answer is 80 12. A constant torque brings a solid, cylinder grinding wheel from rest to an angular speed of 209 rad/s. The wheel has a mass of 1.8 kg and a radius of 0.13 m. What torque in N m is needed to bring the wheel up to that speed in 2.9 s? answer is 1.1
v2
Circular motion: ac =
R
g= 9.8 m/s²;
Pavr =
At
Fnet = ma;
Weight: Fg= mg, ;
1
m v²; Potential energy: Ug = mgy
2
Kinetic energy: K =
E = Er
E = U+K
v?
a, =
= o'r;
Rotational motion:
1 rev = 2n rad;
V = o r;
at = a r;
0 = 00 + at;
0 = 0o t+
2θα-ω-02
2
T=rx F;
T= rFsing;
Στ- Ια;
Ipoint mass = mr?
Idisk =
mR
2
Ipipe
+ Rin
2
1
Irod(center) =
mL²
Laa -mR?
2
Ishell =
3
Thoop = mR?
Irod(end) =-mL?
Iball = -
mR?
dW
P=
dt
I = Icom + MD²
work: W=t 0;
2
W
Pavr =
At
W =
lo 2 +
1
m Vcom 2
2
Rolling:
Vcom = Ro
K =
T= f,R
F,max = H,Fn
Incline: Fer=mgsin0 Fg= mgcos0
Angular momentum:
Lpoint mass =m rxv
L= mrvsin 0;
L=m (r;Vy - ryV3)k
L= Iø
LI = Lr
Ih 01 =I 202
m,X1 +m2X2
X com =
m1y1+m2y2
У com
m1 +m2
m1 +m2
Transcribed Image Text:v2 Circular motion: ac = R g= 9.8 m/s²; Pavr = At Fnet = ma; Weight: Fg= mg, ; 1 m v²; Potential energy: Ug = mgy 2 Kinetic energy: K = E = Er E = U+K v? a, = = o'r; Rotational motion: 1 rev = 2n rad; V = o r; at = a r; 0 = 00 + at; 0 = 0o t+ 2θα-ω-02 2 T=rx F; T= rFsing; Στ- Ια; Ipoint mass = mr? Idisk = mR 2 Ipipe + Rin 2 1 Irod(center) = mL² Laa -mR? 2 Ishell = 3 Thoop = mR? Irod(end) =-mL? Iball = - mR? dW P= dt I = Icom + MD² work: W=t 0; 2 W Pavr = At W = lo 2 + 1 m Vcom 2 2 Rolling: Vcom = Ro K = T= f,R F,max = H,Fn Incline: Fer=mgsin0 Fg= mgcos0 Angular momentum: Lpoint mass =m rxv L= mrvsin 0; L=m (r;Vy - ryV3)k L= Iø LI = Lr Ih 01 =I 202 m,X1 +m2X2 X com = m1y1+m2y2 У com m1 +m2 m1 +m2
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer