<p>A roller coaster car is released from rest at the top of the first rise then moves freely with negligible friction. The roller coaster has a circular loop of radius R in a vertical plane. (height = 2R)</p>
<p>What speed must the car have as it enters the loop to have the minimum speed sqrt(gR) at the top of the loop?</p>
<p>Find the required height of the release point above the bottom of the loop in terms of R.</p>
<p>This one’s really stumping me! Thanks guys for your help.</p>
<p>For the first part, you need to do mgh+.5mv^2 (where h=2R and v=sqrt(gR))=.5mv^2 (where v=v as car enters loop).</p>
<p>For the second part, you do mgh (where h=h of release point)=.5mv^2 (where v=velocity at the bottom of the loop, which you found in the first part).</p>
<p>Does v = (5/2)mgR for the first part?
I’m a little confused how v = .5mv^2</p>
<p>bump. can anyone clarify?</p>
<p>The total energy at the bottom of the loop is: 1/2 mv^2; the total at the top of the loop is: 1/2 m(gR) + mg(2R) = 5/2 mgR. Set these equal to get v = sqrt(5gR). You’ll want to do something similar for the second part.</p>
<p>Thanks! You both were super helpful!</p>