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IV
System of Particles
System of Particles
Rigid body
;an idealization of a solid body of finite
size in which deformation is neglected.
In other words, the distance between
any two given points of a rigid body
remains constant in time regardless of
external forces exerted on it.
System
;a set of interacting or interdependent
components forming an integrated
whole.
Types of force categorised by the source
i. Internal force
ii. External force
𝐹13 = 𝑓𝑜𝑟𝑐𝑒 𝑡ℎ𝑎𝑡 𝑚𝑎𝑠𝑠 3 𝑎𝑐𝑡𝑠 𝑜𝑛 𝑚𝑎𝑠𝑠 1
Forces that mass in the system interacts
with one another are internal force
𝐹13 = − 𝐹31
𝐹𝑡𝑜𝑡𝑎𝑙 = 𝐹𝑒𝑥𝑡
Centre of Mass, 𝐶𝑀
; the weighted average location of all the
mass in a body or group of bodies.
Taking moments at the pivot (origin),
𝑀𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 = 𝑀 𝑎𝑛𝑡𝑖−𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑚1 𝑥1 + 𝑚2 𝑥2 = 𝑚1 + 𝑚2 𝑥 𝐶𝑀
𝑥 𝐶𝑀 =
𝑚1 𝑥1 + 𝑚2 𝑥2
𝑚1 + 𝑚2
∴ 𝑥 𝐶𝑀 =
𝑚𝑖 𝑥𝑖
𝑛
𝑖=1
𝑚𝑖
𝑛
𝑖=1
Suppose the number of mass is infinite,
𝑥 𝐶𝑀 =
lim
𝑛→∞
𝑚𝑖 𝑥𝑖
𝑛
𝑖=1
lim
𝑛→∞
𝑚𝑖
𝑛
𝑖=1
𝑥 𝐶𝑀 =
𝑥𝑖 𝑑𝑚
𝑑𝑚
In 3 dimension,
𝑥 𝐶𝑀 =
𝑥𝑖 𝑑𝑚
𝑑𝑚
𝑦 𝐶𝑀 =
𝑦𝑖 𝑑𝑚
𝑑𝑚
𝑧 𝐶𝑀 =
𝑧𝑖 𝑑𝑚
𝑑𝑚
1
32
𝐹13
𝐹31
𝐹23 𝐹32
𝐹21
𝐹12
𝐹1
𝐹2
𝐹3
𝑦
𝑥
𝑥1
𝐶𝑀
𝑥2
𝑥 𝐶𝑀
0
𝑚2
𝑚1
Since 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘,
***In order to integrate 𝑟 𝑑𝑚, we must
find 𝑑𝑚 in term of 𝑑𝑟. Or, to integrate
𝑥 𝑑𝑚, we must find 𝑑𝑚 in term of 𝑑𝑥.
***All integration in this chapter is
definite integral (from 0 to the end of
the edge) so as to include all masses
through out the object in that particular
axis.
Concept of infinitesimal method
1). Point mass
2). Linear mass (1D)
3). Surface mass (2D)
4). Volumetric mass (3D)
***For 2, 3, and 4, there will be a given
function called
linear/surface/volumetric density.
λ = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑑𝑚
𝑑𝑥
𝑘𝑔
𝑚
𝜎 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑑𝑚
𝑑𝐴
𝑘𝑔
𝑚2
𝜌 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑑𝑚
𝑑𝑉
𝑘𝑔
𝑚3
***In case of 𝜎 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑑𝑚
𝑑𝐴
, we must find 𝑑𝐴 in term of 𝑑𝑥 and
then find 𝑑𝑚 in term of λ and 𝑑𝑥.
𝑟𝐶𝑀 =
𝑟 𝑑𝑚
𝑑𝑚
Motion of CM
Since, 𝑟𝐶𝑀 =
𝑚 𝑖 𝑟 𝑖
𝑛
𝑖=1
𝑚 𝑖
𝑛
𝑖=1
=
𝑚 𝑖 𝑟 𝑖
𝑛
𝑖=1
𝑀 𝑡𝑜𝑡𝑎𝑙
𝑑𝑟𝐶𝑀
𝑑𝑡
=
𝑑
𝑑𝑡
𝑚𝑖 𝑟𝑖
𝑛
𝑖=1
𝑀𝑡𝑜𝑡𝑎𝑙
=
𝑚𝑖
𝑑𝑟𝑖
𝑑𝑡
𝑛
𝑖=1
𝑀𝑡𝑜𝑡𝑎𝑙
𝑣 𝐶𝑀 =
𝑚𝑖 𝑣𝑖
𝑀𝑡𝑜𝑡𝑎𝑙
Find 𝑎 𝐶𝑀,
𝑑𝑣 𝐶𝑀
𝑑𝑡
=
𝑑
𝑑𝑡
𝑚𝑖 𝑣𝑖
𝑛
𝑖=1
𝑀𝑡𝑜𝑡𝑎𝑙
=
𝑚𝑖
𝑑𝑣𝑖
𝑑𝑡
𝑛
𝑖=1
𝑀𝑡𝑜𝑡𝑎𝑙
𝑎 𝐶𝑀 =
𝑚𝑖 𝑎𝑖
𝑀𝑡𝑜𝑡𝑎𝑙
𝑣 𝐶𝑀 =
𝑚𝑖 𝑣𝑖
𝑀𝑡𝑜𝑡𝑎𝑙
𝑎 𝐶𝑀 =
𝑚𝑖 𝑎𝑖
𝑀𝑡𝑜𝑡𝑎𝑙

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4). system of particles (finished)

  • 2. System of Particles Rigid body ;an idealization of a solid body of finite size in which deformation is neglected. In other words, the distance between any two given points of a rigid body remains constant in time regardless of external forces exerted on it. System ;a set of interacting or interdependent components forming an integrated whole. Types of force categorised by the source i. Internal force ii. External force 𝐹13 = 𝑓𝑜𝑟𝑐𝑒 𝑡ℎ𝑎𝑡 𝑚𝑎𝑠𝑠 3 𝑎𝑐𝑡𝑠 𝑜𝑛 𝑚𝑎𝑠𝑠 1 Forces that mass in the system interacts with one another are internal force 𝐹13 = − 𝐹31 𝐹𝑡𝑜𝑡𝑎𝑙 = 𝐹𝑒𝑥𝑡 Centre of Mass, 𝐶𝑀 ; the weighted average location of all the mass in a body or group of bodies. Taking moments at the pivot (origin), 𝑀𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 = 𝑀 𝑎𝑛𝑡𝑖−𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑚1 𝑥1 + 𝑚2 𝑥2 = 𝑚1 + 𝑚2 𝑥 𝐶𝑀 𝑥 𝐶𝑀 = 𝑚1 𝑥1 + 𝑚2 𝑥2 𝑚1 + 𝑚2 ∴ 𝑥 𝐶𝑀 = 𝑚𝑖 𝑥𝑖 𝑛 𝑖=1 𝑚𝑖 𝑛 𝑖=1 Suppose the number of mass is infinite, 𝑥 𝐶𝑀 = lim 𝑛→∞ 𝑚𝑖 𝑥𝑖 𝑛 𝑖=1 lim 𝑛→∞ 𝑚𝑖 𝑛 𝑖=1 𝑥 𝐶𝑀 = 𝑥𝑖 𝑑𝑚 𝑑𝑚 In 3 dimension, 𝑥 𝐶𝑀 = 𝑥𝑖 𝑑𝑚 𝑑𝑚 𝑦 𝐶𝑀 = 𝑦𝑖 𝑑𝑚 𝑑𝑚 𝑧 𝐶𝑀 = 𝑧𝑖 𝑑𝑚 𝑑𝑚 1 32 𝐹13 𝐹31 𝐹23 𝐹32 𝐹21 𝐹12 𝐹1 𝐹2 𝐹3 𝑦 𝑥 𝑥1 𝐶𝑀 𝑥2 𝑥 𝐶𝑀 0 𝑚2 𝑚1
  • 3. Since 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘, ***In order to integrate 𝑟 𝑑𝑚, we must find 𝑑𝑚 in term of 𝑑𝑟. Or, to integrate 𝑥 𝑑𝑚, we must find 𝑑𝑚 in term of 𝑑𝑥. ***All integration in this chapter is definite integral (from 0 to the end of the edge) so as to include all masses through out the object in that particular axis. Concept of infinitesimal method 1). Point mass 2). Linear mass (1D) 3). Surface mass (2D) 4). Volumetric mass (3D) ***For 2, 3, and 4, there will be a given function called linear/surface/volumetric density. λ = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑑𝑚 𝑑𝑥 𝑘𝑔 𝑚 𝜎 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑑𝑚 𝑑𝐴 𝑘𝑔 𝑚2 𝜌 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑑𝑚 𝑑𝑉 𝑘𝑔 𝑚3 ***In case of 𝜎 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑑𝑚 𝑑𝐴 , we must find 𝑑𝐴 in term of 𝑑𝑥 and then find 𝑑𝑚 in term of λ and 𝑑𝑥. 𝑟𝐶𝑀 = 𝑟 𝑑𝑚 𝑑𝑚 Motion of CM Since, 𝑟𝐶𝑀 = 𝑚 𝑖 𝑟 𝑖 𝑛 𝑖=1 𝑚 𝑖 𝑛 𝑖=1 = 𝑚 𝑖 𝑟 𝑖 𝑛 𝑖=1 𝑀 𝑡𝑜𝑡𝑎𝑙 𝑑𝑟𝐶𝑀 𝑑𝑡 = 𝑑 𝑑𝑡 𝑚𝑖 𝑟𝑖 𝑛 𝑖=1 𝑀𝑡𝑜𝑡𝑎𝑙 = 𝑚𝑖 𝑑𝑟𝑖 𝑑𝑡 𝑛 𝑖=1 𝑀𝑡𝑜𝑡𝑎𝑙 𝑣 𝐶𝑀 = 𝑚𝑖 𝑣𝑖 𝑀𝑡𝑜𝑡𝑎𝑙 Find 𝑎 𝐶𝑀, 𝑑𝑣 𝐶𝑀 𝑑𝑡 = 𝑑 𝑑𝑡 𝑚𝑖 𝑣𝑖 𝑛 𝑖=1 𝑀𝑡𝑜𝑡𝑎𝑙 = 𝑚𝑖 𝑑𝑣𝑖 𝑑𝑡 𝑛 𝑖=1 𝑀𝑡𝑜𝑡𝑎𝑙 𝑎 𝐶𝑀 = 𝑚𝑖 𝑎𝑖 𝑀𝑡𝑜𝑡𝑎𝑙 𝑣 𝐶𝑀 = 𝑚𝑖 𝑣𝑖 𝑀𝑡𝑜𝑡𝑎𝑙 𝑎 𝐶𝑀 = 𝑚𝑖 𝑎𝑖 𝑀𝑡𝑜𝑡𝑎𝑙