this is about center of mass, center of mass for complicated shapes, center of mass of hemisphere, center of mass of many particles, center of mass of solids, center of mass of uniform cylinder, center of mass of uniform rod

2. Center Of Mass Of Solids
Definition:
“The point of an
object at which all
the mass of an
object is thought
to be
concentrated is
called center of
mass.”

3. As Balancing Point
The centre of mass of an object is the point at which the
object can be balanced.

4. For Simple Geometric Shapes:
For simple rigid objects with uniform density,
the center of mass is located at the centroid.

5. Center Of Mass For Complicated Shapes:
With Plumb Line

6. EFFECTS OF EXTERNAL FORCES:
Clockwise rotation Anticlockwise rotation
Straight line motion

7. Center Of Mass Of Two Particles:
RCM=m1
r1
+m2
r2
𝑚1
+
𝑚2
For x-co-ordinate:
XCM=
𝑚1𝑥1+𝑚2𝑥2
𝑚1
+
𝑚2

8. Center Of Mass Of Many Particles:
RCM=m1
r1
+m2
r2
+⋯..+mnrn
𝑀
= 𝑖
miri
𝑀
Where M=m1+m2+…+mn
XCM=
𝑚1𝑥1+𝑚2𝑥2+⋯+𝑚𝑛
𝑀
= 𝑖
mixi
𝑀
Similarly for y & z coordinates:
YCM= 𝑖
miyi
𝑀
ZCM= 𝑖
mi
zi
𝑀

9. VELOCITY OF CENTER OF MASS:
AS we know:
RCM=m1
r1
+m2
r2
+⋯..+mnrn
𝑀
Differentiating w.r.t “t” :
𝑑𝑟𝑐𝑚/𝑑𝑡 =
𝑑
𝑑𝑡
(m1
r1
+m2
r2
+⋯..+mnrn
𝑀
)
Vcm=m1
v1
+m2
v2
+⋯..+mnvn
𝑀
= 𝑖
mi
vi
𝑀

10. ACCELERATION OF CM:
Vcm=m1
v1
+m2
v2
+⋯..+mnvn
𝑀
= 𝑖
mi
vi
𝑀
Differentiating again w.r.t. “t”
dVcm/dt =
𝑑
𝑑𝑡
m1
v1
+m2
v2
+⋯..+mnvn
𝑀
acm= m1
a1
+m2
a2
+⋯..+mnan
𝑀

11. Linear Momentum For Number Of Particles :
 P=mv
 For many particles:
 P=P1+P2+P3+……+Pn
 Pcm= i=1
n
mivi………………i
 As we know,
 Vcm= 𝑖
mi
vi
𝑀
 VcmM=
𝑖
mivi
……..j
 Comparing eq i & j:
 Pcm= VcmM

12. Center Of Mass Using Integral Calculus:
RCM= 𝑖
mi
ri
𝑀
XCM≈ 𝑖
∆mixi
𝑀
𝑋CM= lim
∆mi
→
0
𝑖
∆mixi
𝑀
 =
1
𝑀
𝑥𝑑𝑚
Likewise, for yCM and zCM
yCM=
1
𝑀
𝑦𝑑𝑚 & zCM =
1
𝑀
𝑧𝑑𝑚
rcm=
𝟏
𝑴
𝒓𝒅𝒎

13. CENTER OF MASS OF UNIFORM ROD:
rcm=
1
𝑀
𝒓𝑑𝑚
for x-coordinate:
𝑋CM=
1
𝑀
𝑥𝑑𝑚
𝑋CM=
1
𝑀 0
𝑙
𝑥𝑑𝑚 …………eq 2
As we know that mass per unit
length:
λ =
m
L
for small change:
λ =
dm
dL

14. Continued…
λ =
dm
dx
 dm= λdx……………. eq 3
putting eq 3 in eq 1:
𝑋CM=
1
𝑀 0
𝑙
𝑥 λdx
 =
λ
M 0
𝑙
𝑥 dx
 =(M/L)/M . l2/2
 = l2/2l
𝑿CM =L/2
Hence center of mass of uniform rod is L/2.

15. CENTER OF MASS OF UNIFORM Cylinder
rcm=
1
𝑀
𝒓𝑑𝑚
𝑋CM=
1
𝑀 0
𝑙
𝑥𝑑𝑚 ………….eq 1
As we know that mass per
volume is:
𝜌 = 𝑚/𝑣
For small patch/change:
𝜌 = 𝑑𝑚/𝑑𝑣

16. Continued…
dm= 𝜌 dv
dm = 𝜌 𝜋𝑟2 dx…………..eq 2
By Integrating
M=𝜌 𝜋𝑟2
0
𝑙
𝑑𝑥
M =𝜌 𝜋𝑟2l ……...........eq 3
Putting eq 2 & 3 in eq 1:
=
𝜌 𝜋𝑟2
0
𝑙
𝑥 𝑑𝑥
𝑙𝜌 𝜋𝑟2
= L2/2L
=L/2
SO center of mass of uniform cylinder is L/2.

17. CENTER OF MASS OF HEMISPHERE:
Mass of hemisphere =M
Radius =R
For small patch/disc:
Mass =dm
Radius =r
By Pythagoras theorem:
𝑅2=r2+y2

18. Continued……
r2=𝑅2-y2 …………Eq 1
rcm=
1
𝑀
𝒓𝑑𝑚
𝑦𝑐𝑚=
1
𝑀 0
𝑅
𝑦𝑑𝑚
𝜌 = 𝑚/𝑣
𝜌 = 𝑑𝑚/𝑑𝑣
dm= 𝜌𝑑𝑣

19. Continued……
dv= 𝜋𝑟2 dy
dm=𝜌 𝜋𝑟2 dy …………Eq 3
 0
𝑚
𝑑𝑚 = 0
𝑅
𝜌 𝜋𝑟2dy
M=𝜌 𝜋 0
𝑅
𝑟2dy
From eq 1 r2=𝑅2-y2
M= 𝜌 𝜋 0
𝑅
𝑅2dy- 0
𝑅
y2dy
M= 𝜌 𝜋
2
3
𝑅3

20. Continued……
𝑦𝑐𝑚 =
1
𝜌 𝜋
2
3
𝑅3 0
𝑅
𝑦 𝜌 𝜋𝑟2 dy
𝑦𝑐𝑚 =
3
2𝑅3 0
𝑅
𝑦 𝑟2 dy
𝑦𝑐𝑚 =
3
2𝑅3 0
𝑅
𝑦(𝑅2 − y2)
dy
𝑦𝑐𝑚 =
𝟑𝑹
𝟖
So center of mass of hemisphere is
𝟑𝑹
𝟖