Question

At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude of 15.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant and (1/8) revolution later.

Answer 1

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

Step-by-step explanation:

We apply the equations of circular motion uniformly accelerated :

`(a_(T)) ^(2) = (a_(n) )^(2) +(a_(t) )^(2)` Formula (1)

`a_(n) = (v^(2) )/(r)` Formula (2)

`a_(t) = \alpha *r` Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ Formula (5)

Where:

`a_(T)` : total acceleration, (m/s²)

`a_(n)` : normal acceleration, (m/s²)

`a_(t)` : tangential acceleration, (m/s²)

`\alpha` : 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:

`a_(T)` = 15 m/s²

`a_(t)` = 12 m/s²

r=7.90 m :radius of the circular path

Problem development

In the formula (1) :

`a_(n) = \sqrt{(a_(T))^(2) -(a_(t))^(2) }`

We replace the data

`a_(n) = \sqrt{(15)^(2) -(12)^(2)}`

`a_(n) = 9 (m)/(s^(2) )`

We use formula (2) to calculate v:

`9 = (v^(2) )/(7.9)` Equation (1)

a)Speed of the particle at this instant

in the equation (1):

`v=√(9*7.9) = 8.43 (m)/(s)`

b)Speed of the particle at (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

`a_(t)` = 12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

`12 = \alpha *7.90`

`\alpha =(12)/(7.9) = 1.52  (rad)/(s^(2) )`

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