A loop-the-loop has radius 1.5 m; given the minimum top speed 3.83 m/s to maintain contact, what is the minimum speed at the bottom, ignoring losses?

A loop-the-loop has radius 1.5 m; given the minimum top speed 3.83 m/s to maintain contact, what is the minimum speed at the bottom, ignoring losses?

• A. 5.0 m/s
• B. 12.0 m/s
• C. 15.0 m/s
• D. 8.6 m/s

Answer: (D)

At the top, to just stay in contact with the track, the normal force can be zero, so gravity must provide all the centripetal acceleration: v_top^2 / r = g. This gives the minimum top speed v_top = sqrt(g r). With r = 1.5 m, v_top ≈ sqrt(9.8 × 1.5) ≈ 3.83 m/s. From bottom to top, the height increases by 2r, so the gravitational potential energy increases by m g (2r). If there are no losses, energy is conserved, so (1/2) m v_bottom^2 = (1/2) m v_top^2 + m g (2r). This yields v_bottom^2 = v_top^2 + 4 g r. Since v_top^2 = g r, you get v_bottom^2 = 5 g r. Plugging in numbers: 5 g r = 5 × 9.8 × 1.5 = 73.5, so v_bottom ≈ sqrt(73.5) ≈ 8.6 m/s. So the minimum speed at the bottom is about 8.6 m/s.

At the top, to just stay in contact with the track, the normal force can be zero, so gravity must provide all the centripetal acceleration: v_top^2 / r = g. This gives the minimum top speed v_top = sqrt(g r). With r = 1.5 m, v_top ≈ sqrt(9.8 × 1.5) ≈ 3.83 m/s.

From bottom to top, the height increases by 2r, so the gravitational potential energy increases by m g (2r). If there are no losses, energy is conserved, so (1/2) m v_bottom^2 = (1/2) m v_top^2 + m g (2r). This yields v_bottom^2 = v_top^2 + 4 g r. Since v_top^2 = g r, you get v_bottom^2 = 5 g r.

Plugging in numbers: 5 g r = 5 × 9.8 × 1.5 = 73.5, so v_bottom ≈ sqrt(73.5) ≈ 8.6 m/s.

So the minimum speed at the bottom is about 8.6 m/s.