What is the centripetal acceleration of the hammer?

Question

An athlete whirls an 8.96 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 1.67 rev/s. Assume his arm length is included in the length given for the chain. Answer in units of m/s2.

Answer 1

Recall that a_c=v_t²/r where a_c is centripetal acceleration (m/s²), v_t is tangential velocity (m/s) and r is the radius (m) of the "orbit".  We're given the orbital frequency f=1.67 rev/s.  So we need to convert this to v_t.  To do this, we need to figure out what the circumference of the orbit is--that is, how far the hammer travels in one revolution.  Recall that circ= 2πr where π=3.14 and r is the radius of the orbit (m).  So, v_t=f*2πr.  Finally, subbing this into the first equation, we get:  a_c=(f*2πr)²/r = 85m/s²