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.

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 :

Formula (1)

Formula (2)

Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ  Formula (5)

Where:

:  total acceleration, (m/s²)

: normal acceleration, (m/s²)

:  tangential acceleration, (m/s²)

: 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:

= 15 m/s²

=  12 m/s²

r=7.90 m  :radius of the circular path

Problem development

In the formula (1) :

We replace the data

We use formula (2)  to calculate v:

 Equation (1)

a)Speed of the particle at this instant

in the equation (1):

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

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

=  12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

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

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

answer: 31, 5556. 952 seconds