Given Information:
length of slender rod = L = 90 cm = 0.90 m
mass of slender rod = m = 0.120 kg
mass of sphere welded to one end = m₁ = 0.0200 kg
mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)
Required Information:
Linear speed of the 0.0500 kg sphere = v = ?
Answer:
Linear speed of the 0.0500 kg sphere = 1.55 m/s
Explanation:
The velocity of the sphere can by calculated using
ΔKE = ½Iω²
Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy
The moment of inertia of a rigid rod about center is given by
I = (1/12)mL²
The moment of inertia due to m₁ and m₂ is
I = (m₁+m₂)(L/2)²
L/2 means that the spheres are welded at both ends of slender rod whose length is L.
The overall moment of inertia becomes
I = (1/12)mL² + (m₁+m₂)(L/2)²
I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²
I = 0.0081 + 0.01417
I = 0.02227 kg.m²
The change in the potential energy is given by
ΔPE = m₁gh₁ + m₂gh₂
Where h₁ and h₂ are half of the length of slender rod
L/2 = 0.90/2 = 0.45 m
ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45
The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.
ΔPE = -0.1323 J
This potential energy is then converted into kinetic energy therefore,
ΔKE = ½Iω²
0.1323 = ½(0.02227)ω²
ω² = (2*0.1323)/0.02227
ω = √(2*0.1323)/0.02227
ω = 3.45 rad/s
The linear speed is
v = (L/2)ω
v = (0.90/2)*3.45
v = 1.55 m/s
Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.
