# An amusement park ride consists of a rotating circular platform 8.08 m

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Similar to the other rotational problem, the vertical component of tension balances the weight and the horizontal component provides centripetal acceleration.  $Tcos\theta=mg$ and $Tsin\theta=mv^2/r$, which gives $v^2=grtan\theta$.
Here, need to draw a nice diagram to see that $r=lsin\theta+\frac d2$.
With $v$ found, using $Tcos\theta=mg$ will give the tension.