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When a particle moves in a circle, it is constantly changing direction, which means it is accelerating even if its speed is constant. This centripetal acceleration is directed towards the centre of the circle and requires a centripetal force.
If a particle moves in a circle of radius r, its angular speed omega is:
omega = d theta/dt
where theta is the angle swept out. The SI unit is rad s^(-1).
For one complete revolution: theta = 2pi, time = T (period).
omega = 2pi/T = 2pi f
where f is the frequency (revolutions per second).
The linear speed v (tangential speed) is related to angular speed by:
v = r omega
A particle moving at constant speed v in a circle of radius r has acceleration directed towards the centre:
a = v^2/r = r omega^2
This is the centripetal acceleration. It changes the direction of the velocity but not its magnitude.
By Newton's Second Law, the force required to maintain circular motion is:
F = mv^2/r = mr omega^2
This is not a new type of force -- it is the resultant of all forces acting on the particle, directed towards the centre. The centripetal force might be provided by tension, gravity, friction, normal reaction, or a combination.
A particle of mass 0.5 kg moves in a horizontal circle of radius 2 m at 3 rad s^(-1). Find the centripetal force.
Solution:
F = mr omega^2 = 0.5 * 2 * 9 = 9 N
A car travels around a circular bend of radius 50 m at 20 m s^(-1). Find the centripetal acceleration.
Solution:
a = v^2/r = 400/50 = 8 m s^(-2)
A satellite orbits Earth at height h above the surface. The gravitational force provides the centripetal force. Show that v = sqrt(gR^2/(R + h)) where R is Earth's radius.
Solution:
Gravitational force at height h: F = GMm/(R+h)^2 = mg R^2/(R+h)^2
Centripetal force: F = mv^2/(R+h)
Setting equal: mg R^2/(R+h)^2 = mv^2/(R+h)
v^2 = gR^2/(R+h)
v = sqrt(gR^2/(R + h))
Period T = time for one revolution = 2pi/omega = 2pi r/v
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