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3 years ago

What centripetal force is needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s

Physics

1 answer:

liq [111]3 years ago

7 0

Answer:

We conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

Explanation:

Given

Mass m = 7 kg

Radius r = 4 m

Velocity v = 15m/s

To determine

We need to determine the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s.

We know a centripetal force acts on a body to keep it moving along a curved path.

We can determine the centripetal force using the formula

$F_c=\frac{mv^2}{r}$

where

$m$ is the mass

$v$ is the velocity

$r$ is the radius

$F_c$ is the centripetal force

substitute m = 7, r = 4, and v = 15 in the formula

$F_c=\frac{mv^2}{r}$

$F_c=\frac{7\left(15\right)^2}{4}$

$F_c=\frac{1575}{4}$

$F_c=393.75$ N

Therefore, we conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

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At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude o

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Answer:

a) Speed of the particle at this instant

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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 :

$(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2}$ Formula (1)

$a_{n} = \frac{v^{2} }{r}$ 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

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

$a_{n} = 9 \frac{m}{s^{2} }$

We use formula (2) to calculate v:

$9 = \frac{v^{2} }{7.9}$ Equation (1)

a)Speed of the particle at this instant

in the equation (1):

$v=\sqrt{9*7.9} = 8.43 \frac{m}{s}$

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

$12 = \alpha *7.90$

$\alpha =\frac{12}{7.9} = 1.52 \frac{rad}{s^{2} }$

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

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