Answer:
a) Speed of the particle at this instant
v = 8.43 m/s
b) Speed of the particle at (1/8) revolution later
v = 14.83 m/s
Explanation:
We apply the equations of circular motion uniformly accelerated :
[tex](a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2}[/tex] Formula (1)
[tex]a_{n} = \frac{v^{2} }{r}[/tex] Formula (2)
[tex]a_{t} = \alpha *r[/tex] Formula (3)
v= ω*r Formula (4)
ω² = ω₀² + 2*α*θ Formula (5)
Where:
[tex]a_{T}[/tex] : total acceleration, (m/s²)
[tex]a_{n}[/tex] : normal acceleration, (m/s²)
[tex]a_{t}[/tex] : tangential acceleration, (m/s²)
[tex]\alpha[/tex] : angular acceleration (rad/s²)
r : radius of the circular path (m)
v : tangential velocity (m/s)
ω : angular speed ( rad/s)
ω₀: initial angular speed ( rad/s)
θ : angle that the particle travels (rad)
Data:
[tex]a_{T}[/tex] = 15 m/s²
[tex]a_{t}[/tex] = 12 m/s²
r=7.90 m :radius of the circular path
Problem development
In the formula (1) :
[tex]a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }[/tex]
We replace the data
[tex]a_{n} = \sqrt{(15)^{2} -(12)^{2}}[/tex]
[tex]a_{n} = 9 \frac{m}{s^{2} }[/tex]
We use formula (2) to calculate v:
[tex]9 = \frac{v^{2} }{7.9}[/tex] Equation (1)
a)Speed of the particle at this instant
in the equation (1):
[tex]v=\sqrt{9*7.9} = 8.43 \frac{m}{s}[/tex]
b)Speed of the particle at (1/8) revolution later
We know the following data:
θ =(1/8) revolution=( 1/8) *2π= π/4
[tex]a_{t}[/tex] = 12 m/s²
v₀= 8.43 m/s
r=7.9 m
We use formula (3) to calculate α
[tex] 12 = \alpha *7.90[/tex]
[tex]\alpha =\frac{12}{7.9} = 1.52 \frac{rad}{s^{2} }[/tex]
We use formula (4) to calculate ω₀
v₀= ω₀ *r
8.43 = ω₀*7.9
ω₀ = 8.43/7.9 = 1.067 rad/s
We use formula (5) to calculate ω
ω² = ω₀² + 2*α*θ
ω²= (1.067)² + 2*1.52*π/4
ω² =3.526
ω = 1.87 rad/s
We use formula (4) to calculate v
v= 1.87 rad/s * 7.9m
v = 14.83 m/s : speed of the particle at (1/8) revolution later
