Question

An athlete whirls a 6.86kg hammer tied to the end of a 1.3m chain im a simple horizontal circle.Where you should ignore any vertical deviations.The hammer moves at the rate of 0.844 rev/s. What is the centripetal acceleration of the hammer?

Answer 1

Answer on Question 83678, Physics, Mechanics, Relativity

Question:

An athlete whirls a 6.86 kg hammer tied to the end of a 1.3 m chain in a simple horizontal circle. Where you should ignore any vertical deviations. The hammer moves at the rate of 0.844 rev/s. What is the centripetal acceleration of the hammer?

Solution:

The centripetal acceleration of the hammer can be found from the formula:

$$a_c=v^2/r,$$

here,

$$v$$

is the linear velocity of the hammer,

$$r=1.3 m$$

is the radius of the circular path.

From the other hand, the relationship between the linear velocity

$$v$$

and angular velocity

$$ω$$

of the rotating object is given by the formula:

$$v=ωr.$$

Finally, substituting this formula into the first equation, we get:

$$a_c=v^2/r=(ωr)^2/r=ω^2r.$$

Let's substitute the numbers:

$$a_c=(0.844 rev/s * 2π rad/1 rev)^2*1.3 m=36.5 m/s^2.$$

Answer:

$$a_c=36.5 m/s^2$$

.