Question

a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does the child experience?(unit=m/s^2)

Answer 1

Answer: $0.146 m/s^{2}$

Explanation:

The centripetal acceleration $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$  (1)

Where:

$V$ is the tangential velocity

$r=1.05 m$ is the radius of the circle

On the other hand, the tangential velocity  is expressed as:

$V=\omega r$ (2)

Where $\omega$ is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

$\omega=\frac{5 rev}{13.4 s}=0.37 rev/s$ (3)

Substituting (3) in (2):

$V=(0.37 rev/s)(1.05 m)$ (4)

$V=0.39 m/s$ (5)

Substituting (5) in (1):

$a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}$  (6)

Finally:

$a_{c}=0.146 m/s^{2}$  

Answer 2

Answer:

The child will experience a centripetal acceleration of  0.15 m/$s^{2}$

Explanation:

Centripetal acceleration is the acceleration of the object along a circular path which tends to move towards the center of the circular path.

The Centripetal acceleration of the child can be obtained with the expression below (equation I);

$a_{c} = \frac{v^{2} }{r}$                                           I

where $a_{c}$ is the centripetal acceleration

           v is the tangential velocity

           and r is the radius of the path.

the tangential velocity of the body can be gotten using the equation below (equation II);

V = ωr                                                              II

where v is the tangential velocity

           ω is the angular velocity

           r, of course, is our radius

Now ω is expressed as  the rate of change of angular position with time;

ω = $\frac{change in angular distance}{time}$               III

from the question  change in angular distance is 5 rev and time is 13.4 s, substituting in equation III we have;

ω  =$\frac{5rev}{13.4s}$

ω = 0.373 rev/s

Now angular velocity had been obtain, we substitute into equation II to get tangential Velocity;

V = ?

ω = 0.373 rev/s

r = 1.05 m

V = 0.373 rev/s × 1.05 m

V = 0.39165 m/s.

the tangential velocity is 0.39165 m/s, this will be substituted in equation I to get centripetal acceleration of the child.

From the question;

$a_{c}$ = ?

v = 0.39165 m/s

r =  1.05 m

using equation I;

$a_{c} = \frac{0.39165 m/s^{2} }{1.05 m}$

$a_{c}$ = 0.15 m/$s^{2}$

Therefore the centripetal acceleration that the child will experience is 0.15 m/$s^{2}$