Question

A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children swings from this rope that is 6.00 m long, his tangential speed at the bottom of the swing is 9.50 m/s. What is the centripetal acceleration, in m/s2, of the child at the bottom of the swing

Answer 1

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

The centripetal acceleration is given by:

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

Where:

$v^{2}$: is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

Answer 2

Answer:

10.8n

Explanation:

For this case the centripetal force is given by:

Where,

m: mass of the object

v: tangential speed

r: rope radius

Substituting values in the equation we have:

Then, doing the corresponding calculations:

Answer:

The centripetal force exerted on the rope is: