Centripetal force is the inward force that acts on an object moving along a circular path and keeps it confined to that path. This force continuously changes the direction of the object’s velocity without necessarily changing its speed. Since velocity is a vector quantity, any change in its direction results in acceleration, known as centripetal acceleration. The magnitude of this force depends on the speed of the object and the radius of the circular path—higher speed or a smaller radius requires a greater centripetal force to maintain circular motion.
\boxed {a_c = \frac{v^2}{r}}
where,
- ac is centripetal acceleration
- v is velocity of object moving in circular path
- r is radius of circular path
In circular motion, the direction of velocity changes continuously, which results in acceleration. This acceleration always points toward the center of the circular path and is called centripetal acceleration (ac ). The term centripetal means toward the center.
Derivation of Centripetal Acceleration Formula
The figure given below shows an object which is moving in a circular motion. Assume that the speed of the object is constant. The figure shows the direction of the instantaneous velocity and points B and C. Since acceleration is the change in velocity, the change in velocity is roughly pointing towards the center of the curvature. To measure things instantaneously, the points B and C must be brought really close together, and the angle between them should be infinitely small. In that case, the change of velocity will point directly towards the center of the curvature.
In the figure given above, notice that the triangles ABC and QPR are isosceles triangles. The two sides of the vector velocity triangles are denoted by v1 and v2. Comparing the two triangles using properties of similar triangles,
The acceleration is given by the change in velocity with respect to time,
Dividing both sides with delta-t
Rearranging the equation,
Note that Δv/Δt is the ac and Δs/Δt is the tangential speed (v).
So, the centripetal acceleration becomes,
This gives us the acceleration of an object under the circular motion traveling at a speed "v" and with radius "r". This equation depends on the square of the velocity and inversely to the radius "r".
Characteristics
The characteristic of centripetal acceleration is given below:
- Direction: Centripetal acceleration always points toward the center of the circular path. It is perpendicular to the velocity of the object and acts along the radius inwards.
- Magnitude: The magnitude of centripetal acceleration is directly proportional to the square of the object’s speed (v2) and inversely proportional to the radius of the circular path (r)
a_c \propto \frac{v^2}{r} . - Relation to Force: According to Newton’s second law (F= ma), a centripetal acceleration requires a force acting on the object. This centripetal force is always directed toward the center of the circle and is necessary to maintain the circular motion.
Centripetal Force
Centripetal force is the inward force that keeps an object moving along a circular path by continuously changing the direction of its velocity. Without this force the object would move in a straight line. The formula for centripetal force is given as
F_c = m\,a_c
where,
- Fc = Centripetal Force
- m = mass of object
- ac = centripetal acceleration
Applications
Centripetal acceleration plays a crucial role in understanding uniform circular motion, rotational dynamics, and the relationship between force, mass, and acceleration. Its applications include.
- Circular Motion in Nature: It explains the motion of planets around the Sun satellites orbiting celestial bodies and vehicles navigating curved roads or tracks.
- Rides and Amusement Parks: Roller coasters, carousels, and similar rides rely on centripetal acceleration to create thrilling and dynamic experiences for riders.
- Engineering and Design: Engineers consider centripetal acceleration when designing rotating systems such as centrifuges, flywheels, and gyroscopes to ensure proper functioning and safety.
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Solved Problems
Question 1: Find the centripetal acceleration on an object performing circular motion with a radius of 5m. The velocity of the object is 10m/s.
Solution: The centripetal acceleration is given by,
a = \frac{v^2}{r} Given:
v = 10m/s.
r = 5m.
Plugging the values in the equation,
a = \frac{v^2}{r} ⇒ a = (10)2/(5)
⇒ a = 100/5
⇒ a = 20 m/s2
Question 2: An object(m = 5Kg) is performing circular motion with radius 2m. If the velocity of the object is 8 m/s, find the centripetal force acting on the object.
Solution: The centripetal acceleration is given by,
a = \frac{v^2}{r} Given:
v = 8m/s.
r = 2m.
Plugging the values in the equation,
a = \frac{v^2}{r} ⇒ a = (8)2/(2)
⇒ a = 64/2
⇒ a = 32 m/s2
Force acting on the object is given by,
F = ma
⇒ F = (5)(32)
⇒ F = 160 N
Question 3: An object(m = 2Kg) is performing circular motion with radius 5m. If the velocity of the object is 10 m/s, find the centripetal force acting on the object.
Solution: The centripetal acceleration is given by,
a = \frac{v^2}{r} Given:
v = 10m/s.
r = 5m.
Plugging the values in the equation,
a = \frac{v^2}{r} ⇒ a = (10)2/(5)
⇒ a = 100/5
⇒ a = 20 m/s2
Force acting on the object is given by,
F = ma
⇒ F = (2)(20)
⇒ F = 40 N
Question 4: An object(m = 2Kg) is performing circular motion with radius 5m. If the centripetal force acting on the object is 100N, find the velocity of the object.
Solution: Force acting on the object is given by,
F = ma
⇒ 100 = (2)(a)
⇒ a = 50 m/s2
The acceleration of the object is given by,
a = \frac{v^2}{r} ⇒ 50 = v2/5
⇒ 250 = v2
⇒ v = 5√10 m/s.
Unsolved Problems
Question 1: A satellite orbits the Earth in a circular orbit of radius 7×106 m with orbital speed 7.5×103 m/s. Calculate the centripetal acceleration of the satellite.
Question 2: A roller coaster car of mass 200 kg moves along a vertical loop of radius 10 m. Calculate:
a) The centripetal acceleration at the top of the loop if its speed there is 15 m/s,
b) The normal force acting on the car at the top of the loop.
Question 3: An object is performing circular motion on a friction less horizontal table. The radius of the circle is halved while the speed is doubled. How does the centripetal acceleration change?
Question 4: A roller coaster car of mass 500 kg moves along a vertical circular loop of radius 20 m. If the centripetal force at the top of the loop is 4000 N, calculate the speed of the car at that point.
Question 5: Two objects of equal mass move in circular paths of different radii r1 and r2 with the same speed v. Compare the centripetal forces acting on them.