A 60 kg gila monster on a merry go round is travelling in a circle with a radius of 3 m at a speed of 2m/s

Physics

1 answer:

DIA [1.3K]3 years ago

8 0

Acceleration: 1.3 m/s².
Net Force: 80 N (2 sig. fig.).
Weight: 5.9 × 10² N = 590 N.

<h3>Explanation</h3>

Apply the equation for objects that move in circles:

$a = \dfrac{v^2}{r}$ ,

where

$a$ is the acceleration of the object,
$v$ is the tangential speed of the object, and
$r$ is the radius of the circular path.

For this monster:

$a$ is to be found,
$v = 2\;m\cdot s^{-1}$ , and
$r = 3\;\text{m}$ .

$\displaystyle a = \frac{({2\;\text{m}\cdot\text{s}^{-1}})^{2}}{3\;\text{m}} = \frac{2^2}{3}\;\text{m}\cdot\text{s}^{-2}= 1.3\;\text{m}\cdot\text{s}^{-2}$ .

By Newton's Second Law,

$\Sigma F = m\cdot a$ ,

where

$\Sigma F$ is the net force on an object,
$m$ is the mass of the object, and
$a$ is the acceleration of the object.

$\Sigma F$ is to be found,
$m = 60\;\text{kg}$ for this monster, and
$a = 1.33333\;\text{m}\cdot\text{s}^{-2}$ from previous calculations.

$\Sigma F = m\cdot a = 60\;\text{kg} \times 1.33333\;\text{m}\cdot\text{s}^{-2} = 80\;\text{N}$ .

Weight of an object near the surface of the earth:

$W = m\cdot g$ ,

where

$m$ is the mass of the object, and
$g$ is the gravitational acceleration "constant" (a.k.a. gravitational field strength.) $g \approx 9.81\;\text{N}\cdot\text{kg}^{-1}$ near the surface of the earth.