Question

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

Answer 1

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

Explanation

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.

$W = 60\;\text{kg} \times 9.81\;\text{N}\cdot\text{kg}^{-1}=5.9\times 10^{2} \;\text{N}$.