`F_c=(mv^2)/r` Show
Fc is the centripetal force of the circular motion, m is the mass (kg) of the object undergoing the motion, v (m/s) is the linear or tangential velocity of the object and r (m) is the radius of this circular motion. Centripetal force provides the curvature to an object’s circular motion. Its direction is orthogonal (right-angled) to the direction of the object’s motion (velocity). In other words, it is always towards the centre of the circular motion. Centripetal acceleration (which can be calculated by Newton’s second law) is in the same direction as centripetal force. Centripetal force is a non-real forceThis means that centripetal force always caused by a real force. For example:
Relationship between centripetal force, mass, speed and radiusFrom the formula, we can deduce the following:
Effect on centripetal force
Changes in centripetal force
Concept Question 1A car rounds a bend on a road that follows the arc of a circle with radius r. The car has mass m and is travelling at a velocity v. Explain the following situations: (a) Why are drivers advised to slow down during wet weather, specifically when they are making a bend. (b) Assuming the friction between the tyres and the road does not change, describe the path of a car with mass 2m when it rounds the bend at velocity v? (c) A motor cyclist rounds the same bend at velocity 2v. If the mass of the motorcycle is 0.25m, what would be different about the centripetal force acting on the motorcycle compared to that on the car? Concept Question 2HSC Q30 2013 The diagram shows a futuristic space station designed to simulate gravity in a weightless environment.
Concept Questions SolutionsQuestion 1 (a) During wet weather, the kinetic friction between a car's tyres and the ground is reduced. This means the centripetal force acting on the car during its bend is reduced. As a result, velocity needs to decrease to maintain radius of the curvature. (b) Since centripetal force remains constant, the radius of curvature is doubled. This means for a car with mass 2m travelling at the same speed v, it requires a greater distance to complete the bend. (c) As seen above, substituting 0.25m and 2v into the equation `F_c=(mv^2)/r` will yield a magnitude of centripetal force identical to one with mass m and velocity v. This means the centripetal force acting on the motorcycle remains unchanged and so does its radius. Question 2 (a) The rotating motion of the spacecraft exerts normal force (centripetal force) on the astronaut. Due to Newton's third law, the astronaut exerts reaction force on the outer perimeter of the spacecraft. The acceleration resulted from this reaction force simulates gravity. (b) As v increases, the magnitude of centripetal force increases (since radius remains constant). Changes in centripetal force are always proportional to the square of change in velocity. (c) Yes, to maintain gravity, the magnitude of centripetal force cannot be changed. A decrease in radius needs to be compensated by a reduction in velocity v. This means the rotational speed to simulate gravity is lower for smaller space stations. (d) No, the mass of the space station does not affect centripetal force nor acceleration. In addition, the mass of the astronaut does not influence the rotational speed required to achieve 1g of gravitational acceleration. This is because simulated gravity is independent of mass: `a_c=v^2/r`.
In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The centripetal force is responsible for the centripetal motion of the object, and the centrifugal force acting against it prevents the object from meeting the centre of the circle. The centripetal acceleration and radius are obviously related to each other, as the object traveling in the circular track with the application of the centripetal force keeps the acceleration of the object along the radius of the circle. Both the vectors remain in a unique direction. How is centripetal acceleration related to radius?The centripetal acceleration comes into the picture while the object is in a circular motion and elapses the path in a circle having specific radii. The object traveling in circular motion accelerates, maintaining a constant distance from the centre, which is the radius of the circular track. The direction of the centripetal acceleration of the object acts towards the centre of the circular path traced by an object along the radius of the circle. The force that keeps the object in a circular motion is a centripetal force. But this is not only responsible for the object moving along the circular path. The force acts opposing the centripetal force and prevents them from falling inward. The force that keeps them into place while traveling in a circular path is a centrifugal force. Find Centripetal Acceleration from RadiusThe object accelerating in the circular motion exerts a centripetal force. The centripetal force accelerating the object with velocity v is given by the expression: F=mv2/r Here, m is the mass of the object, and r is the radius of the circular path. Since F=ma from Newton’s second law of motion. Using this in the above equation, we get: ma=mv2/r Hence, the formula to find the centripetal acceleration of the object in a circular motion is given as: a=mv2/r According to this equation, the centripetal acceleration of the object is half of the square of the velocity of the object. The linear velocity is always perpendicular to the centripetal acceleration acting inward. The above equation clearly indicates that the angular acceleration is inversely related to the radius of the circular path. This implies that we shall have the highest centripetal acceleration of the object for the object propagating on small radii circles. Centripetal Acceleration and Radius GraphNow, let us understand the inverse relationship between the centripetal acceleration and the radius of the circular trajectory by working on one simple example. Suppose different merry-go-rounds having varying radii are brought in a place to study the effect of the radius of the rotating wheel on the centripetal acceleration of the merry-go-round. The torque is applied on all the merry-go-rounds one by one maintaining the velocity of 3m/s constant. Merry-go-round; Image Credit: pixabayFor the first merry-go-round, the radius r=1 m, hence the centripetal acceleration for this wheel is, a1=mv2/r = The second merry-go-round has a radius r=2 m, hence the centripetal acceleration for this wheel is, a2=v2/r The third merry-go-round has a radius r=3 m, hence the centripetal acceleration for this wheel is, a3=v2/r The fourth merry-go-round has a radius r=4 m, hence the centripetal acceleration for this wheel is, a4=v2/r The fifth merry-go-round has a radius r=5 m, hence the centripetal acceleration for this wheel is, a5=v2/r The data obtained is noted down in a below table:
Let us plot a graph of centripetal acceleration v/s radius for the above data. From the above graph, we can say that the centripetal acceleration of the object in a circular motion decreases exponentially with the increasing radius. The magnitude of the centripetal acceleration decreases along the radius. Hence, to maintain the centripetal acceleration, the velocity of the object has to be increased as the circumference of the path traveled by the object increases. This is due to the fact that, as the radius increases, the centripetal force acting on the object reduces. We can relate this to the Coulomb force between the two unlike charges. As the linear spacing between the two increases, the magnitude of the force is reduced. What happens to centripetal acceleration if radius is doubled?The centripetal acceleration of the object is more for a small radius of the circle as compared to the larger radii. If the radius of the circular path is doubled, keeping the radial velocity of the object constant then, the centripetal acceleration of the object will be reduced to half. If the object is moving with a velocity ‘u’ along a circular path having a radius ‘r’, and the same object is moving on another circular track having a radius ‘2r’ with the velocity ‘v’, then the change in the centripetal acceleration is, This equation gives the change in the centripetal acceleration of the object moving with different velocities while traveling on different circular tracks having a different radius. What is the centripetal acceleration of the car moving along the circular track on the stadium with a velocity of 20 km/h? The diameter of the stadium is 70 meters.Given: The velocity of the car is, The diameter of the stadium is, d= 70 m. Hence, the radius of the stadium is, r= 35 m. The formula to calculate the centripetal acceleration of the car along the stadium is, a=v2/r Substituting the values in this equation, we get, Hence, the centripetal acceleration of a car is 0.86 m/s2 while driving around the stadium. What happens to the centripetal acceleration of the object after reducing the length of the rope to half to which it was attached at the length of 100 cm if the velocity of the object is doubled?Given: The initial length of the rope is, l1=100 cm = 1 m The final length of the rope is, l2=100/2 cm = 50 cm = 0.5 m Let the initial velocity of the object be ‘u’ and the final velocity be ‘v’. The final velocity is doubled the initial velocity, hence, v=2u. The initial centripetal acceleration of the object is, The final centripetal acceleration of the object is, Hence, we can see that the acceleration of the object increases 16 times more than the initial centripetal acceleration of the object after reducing the length of the rope to half. ConclusionThe centripetal acceleration of the object moving in a circular track relies upon the radius of the circle. The centripetal acceleration is directed inward along the radius of the circle. Both are inversely related to each other. If the radius decreases, the centripetal acceleration of the object increases at an exponential rate. |