Motion in a vertical circle
The string slackens at the top if v_top < √(gr); complete loops need v_bottom ≥ √(5gr).
In this lesson
- The string slackens at the top if v_top < √(gr); complete loops need v_bottom ≥ √(5gr).
The roller coaster question: how fast at the bottom to survive the top?
At the top, gravity alone can provide the centripetal force: mg = mv²/r → v_min = √(gr). Energy from bottom to top: ½mv_b² = ½mv_t² + 2mgr → v_b,min = √(5gr).
Key results: T_top = m(v²/r − g), T_bottom = m(v²/r + g). Tension is maximum at the bottom, minimum at the top.
Vertical circle limits
Worked example
In uniform circular motion, the acceleration of the particle is directed:
- Speed is constant but velocity direction changes continuously.
- The change in velocity points toward the centre → centripetal acceleration a_c = v²/r toward the centre.
Answer: Toward the centre of the circle
Transcript (0 min)
WEBVTT 1 00:00:00.000 --> 00:00:05.000 Motion in a vertical circle — FemtoLearn. 2 00:00:05.000 --> 00:00:15.000 The string slackens at the top if v_top < √(gr); complete loops need v_bottom ≥ √(5gr).
Frequently asked
Where is tension maximum?
At the bottom, where gravity and centripetal demand add.
Practice
Free · 4 questions with full solutions- Q1 · Numerical · difficulty 1/5
A wheel rotates at 270 rpm. What is its angular velocity in rad/s?
- Q2 · Numerical · difficulty 1/5
A point is at distance 1.7 m from the axis of a wheel rotating with angular velocity 11 rad/s. Find its linear speed.
- Q3 · Numerical · difficulty 2/5
A particle moves on a circle of radius 1.5 m with constant speed 6 m/s. Find the magnitude of its centripetal acceleration.
- Q4 · MCQ · difficulty 1/5
In uniform circular motion, the acceleration of the particle is directed:
- Along the tangent to the circle
- Toward the centre of the circle
- Away from the centre of the circle
- It is zero