Question

An athlete whirls a 6.55 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 1.4 rev/s. What is the tension in the chain?

Answer in units of N

Answer 1

336.14

Step-by-step explanation:

First we need to find the centripetal acceleration of the hammer.

`a_c=(v^2)/(r)`

We need to find V.

`V=(2\pi r)/(T)`

`(2\pi (1.3))/(1)`

`= 8.168 m/s`

Next we need to use what we calculated and substitute to find the centripetal acceleration.

`a_c=(v^2)/(r) =((8.168)^2)/(1.3) =51.32m/s^2`

Now we are required to find the tension of the chain.

To do so,

`F_t=F_c=m(v^2)/(r)`

or

`F_T=ma_c`

`=(6.55)(51.32)`

`F_T=336.14`

Note: Using the given information, for better understanding you could draw a FBD of the hammer as if it were completely horizontal.