<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter). </span> a=v^2/r v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s. </span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec. </span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s </span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min). (32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min </span></span>