A 0.20 kg stone is whirled in a horizontal circle on a string of length 0.80 m at 5.0 rad s−1. Find the tension in the string that provides the centripetal force.
- Use Fc = mrω2.
- ω2 = 5.02 = 25 rad2 s−2.
- Fc = 0.20 × 0.80 × 25 = 4.0 N.
Answer: 4.0 N
Angular motion and centripetal force
Physics · Circular motion. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
When a body turns about an axis, we describe its motion with angles. Angular displacement θ is the angle turned through, measured in radians. Angular velocity ω is the angle turned per second, and angular acceleration α is the rate of change of ω.
Each point of a rotating body also has linear motion along its circular path. Multiply the angular quantity by the radius to get the linear one.
A body moving in a circle keeps changing direction, so it accelerates towards the centre even at constant speed. The force that causes this is the centripetal force. It is not a new kind of force: friction, tension or gravity provides it. If it disappears, the body moves off along the tangent.
Rules to remember
- angle in radians = arc length ÷ radius (θ = s/r); 1 revolution = 2π rad = 360°, so 1 rad = 57.3°
- angular velocity = angular displacement ÷ time (ω = Δθ/Δt), in rad s−1; for n revolutions per second, ω = 2πn
- angular acceleration = change in angular velocity ÷ time (α = Δω/Δt), in rad s−2
- linear and angular: s = rθ, v = rω, at = rα (θ must be in radians)
- centripetal acceleration = v2/r = rω2, directed towards the centre
- centripetal force = mv2/r = mrω2
- least speed at the top of a vertical circle: weight alone gives the centripetal force, so v = √(gr)
Common mistake
Revolutions per minute are not rad s−1. Divide by 60 to get revolutions per second, then multiply by 2π.
Worked example
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