Circular MotionPart 2

Centripetal Force

Bear in mind Centripetal force is not a new force or a separate force. It is a name given for the resultant of the usual forces such as friction, normal reaction, tension of the string, force of gravity, EM forces etc that causes circular motion. This is the resultant force must be directed towards the centre of the circle.

(b) Centripetal force does not change the magnitude of the velocity; It only changes the direction of the motion.

(c) Direction of F is perpendicular to the direction of the motion. There is no component of the force in the direction of the motion. Therefore, no work is done on the body.

Centripetal ForcePractice 1

Example of circular motion

Practice 2

Example of circular motion

Circling at an angle

Circling at an angle

Example of circular motion

Example of circular motion

Practice 3

Example of circular motion

Example of circular motion

Example of circular motion

Practice 4

Practice 5

Practice 6

Example of circular motion

Practice 7

Example of circular motion

Practice 8

Banking of Roads

The phenomenon of raising outer edge of the curved road above the inner edge is to provide necessary centripetal force to the vehicles to take a safer turn and the curved road is calledBanking of Roads.

Banked RoadsWhen a vehicle goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the vehicle have a tendency to leave the curved path and regain the straight line path. Force of friction between wheels and the roads opposes this tendency of the wheels. This force of friction therefore, acts towards the centre of circular track and provides the necessary centripetal force.

Centripetal Force on Banked RoadsThe centripetal force is along the radius of curvature of the banked road which is horizontal and not along the incline. This is also the direction of the centripetal acceleration.

A car of mass M = 1300 kg traveling at 55.0 km/hour enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires. Use g = 9.80 m/s2throughout this problem.r = 65.4 m

FNsin = mac

Since ac= v2/r: FNsin = mv2/r: And we know (from the explanation above) that FNcos = mg: ((FNsin) / (FNcos)) = (mv2/r) / (mg) Which gives us: tan = (mv2/r) / (mg) simplified to: tan = v2/ (rg) We get the tan because the FN's on the left cancel and sin/cos = tan. So to solve now: First, find velocity in terms of m/s: 55km/h = 55000m/h = 15.277m/s Next use the formula we solved for: tan = v2/ (rg) tan(20) = 15.2772/r) / (9.8r) 0.364 = (23.81/r) r = 65.4m

Non-uniform circular motionis any case in which an object moving in a circular path has a varyingspeed. Some examples of non-uniform circular motion include a roller coaster, a verticalpendulum, and a car riding over a hill. All of these situations include an object traveling at different speeds in a circular path.Non-uniform Circular Motion

Non-uniform Circular MotionChange the speedHere is angular acceleration

Rotational kinematicsFor constant angular acceleration

Differences from uniform circular motion

If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.

This concept can be used for an object moving along any curved path, as any small segment of the path will be approximately circular.

A simple pendulum consists of a mass, m suspended at the end of a rope oflength, l. When the rope is at an angle , to the vertical, the speed of the bob is, v. Find :(a) the radial and tangential components of the acceleration (b) the tension in the threadSOLUTION:It is obvious that the path of the bob is a partof a vertical circle of radius, l. When thread makes an angle, with the vertical, the speed of the bob is given to be, v. At this moment there areonly two forces acting on theball : (1) tension force, T; (2) its weight, mg

From figure (b) it is clear that mg sin is the only tangential force on the bob and hence, tangential acceleration of the bob is g sin .

Both radial and tangential accelerations areabove in (c). Theirresultant, net acceleration,is also shown. You should note that when thebob reaches at theextreme position its speed becomes zero and hence at that position it hasonly tangential acceleration.(b) (c) (a)

*********************Figure 5-25. Caption: The speed of an object moving in a circle changes if the force on it has a tangential component, Ftan. Part (a) shows the force F and its vector components; part (b) shows the acceleration vector and its vector components.

*Figure 5-26. Caption: Object following a curved path (solid line). At point P the path has a radius of curvature r. The object has velocity v, tangential acceleration atan (the object is here increasing in speed), and radial (centripetal) acceleration aR (magnitude aR = v2/r) which points toward the center of curvature C.

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