What is the minimum speed at the top of a vertical loop to maintain contact with the track?

What is the minimum speed at the top of a vertical loop to maintain contact with the track?

• A. sqrt(2 g R)
• B. sqrt(g R)
• C. sqrt(g R/2)
• D. sqrt(3 g R)

Answer: (B)

At the top of the loop, the object needs a centripetal acceleration toward the loop’s center, which equals v^2/R. The forces toward the center are gravity mg and the normal force from the track. For the smallest speed that still keeps contact, the normal force can be zero, so gravity alone provides the required inward pull: mv^2/R = mg. This gives v^2 = gR, so the minimum speed is v = sqrt(gR). If you go faster, the track would have to push inward (N > 0) to provide the extra centripetal force; if you go slower, gravity isn’t enough to supply v^2/R, and you would lose contact before reaching the top.

At the top of the loop, the object needs a centripetal acceleration toward the loop’s center, which equals v^2/R. The forces toward the center are gravity mg and the normal force from the track. For the smallest speed that still keeps contact, the normal force can be zero, so gravity alone provides the required inward pull: mv^2/R = mg. This gives v^2 = gR, so the minimum speed is v = sqrt(gR).

If you go faster, the track would have to push inward (N > 0) to provide the extra centripetal force; if you go slower, gravity isn’t enough to supply v^2/R, and you would lose contact before reaching the top.