Revision of Horizontal Circular Motion The angular velocity = The tangential velocity = Force towards the centre = Time for 1 revolution =  rads s -1  is constant and does not change v ms –1 v is constant and does not change

Motion in a vertical circle is an example of non–uniform circular motion. An object will slow down as it travels in an upward direction round the circle and then speed up in a downward direction. Suppose that an object of mass m is attached to the end of a light rigid rod, or light string, of length r. The other end of the rod, or string, is attached to a stationary pivot in such a manner that the object is free to execute a vertical circle about this pivot.

Motion in a vertical circle is an example of non–uniform circular motion. An object will slow down as it travels in an upward direction round the circle and then speed up in a downward direction. Suppose that an object of mass m is attached to the end of a light rigid rod, or light string, of length r. The other end of the rod, or string, is attached to a stationary pivot in such a manner that the object is free to execute a vertical circle about this pivot. Let  measure the angular position of the object, measured from the downward vertical. Let u be the velocity of the object at  = 0.

Tangential and radial acceleration. With horizontal circles, the speed remained constant so there was no acceleration in its direction of motion. The only acceleration was radial. (towards the centre of the circle) With vertical circles, the particle slows down as it gains height so there will be an acceleration in its direction of motion. This is called the tangential or transverse acceleration. a T will be negative i.e as  increases a T decreases

Formula for tangential acceleration Suppose the particle makes angle  with the downward vertical after t seconds and has a tangential velocity v T and tangential acc a T. l = r  (length of arc from core 2, where  must be in radians) as r is a constant For motion in a horizontal circlev = r  where  is the constant rate of change of  If  varies as in vertical circles then  = = So v T = r where means a T =  vTvT aTaT

v T = r  Variable VelocityConstant Velocity a T = 0

1) If  = 2t 5, r = 0.3m find a T and a r after 2 seconds. = 10t 4 = 40t 3 a T = r = 0.3×40t 3 = 12t 3 = 96ms -2 when t = 2. also, a r = = ×0.3 = 7680ms -2 when t = 2.

2)A 0.2kg mass on the end of a 0.5m string performs vertical circles. At the instant when the string makes a 30° angle with the downward vertical the tension in the string is 3N. Find both components of its acceleration. F = ma towards centre of circle, 3 – 0.2gcos30 = 0.2a r so a r = 6.51 ms -2 F = ma tangentially –0.2gsin30 = 0.2a T so a T = –4.9 ms -2 T=3 30 o aa 0.2g EX D1.1 pg 331–4

PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D The object moves from point A, where its tangential velocity is u, to point B, where its tangential velocity is v. Consider energy conservation. At point A, the object is situated a vertical distance r below the pivot, whereas at point B the vertical distance below the pivot has been reduced to rcos . Hence, in moving from A to B the object gains potential energy and loses kinetic energy as it slows down. Take the zero potential energy line to be the horizontal line through the centre of the circle. Hence the potential energy at A is –mgr, the energy at the D is +mgr and the energy at B is –mgrcos . Consider an object moving around a vertical circle of radius r. rcos  

Using conservation of energy Energy at A = Energy at B  mu 2 – mgr =  mv 2 – mgrcos  which reduces to v 2 = u 2 + 2gr(cos  - 1) PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D rcos  

The radial forces acting on the object are the tension T in the rod, or string, which acts towards the centre of the circle, and the component mgcos  of the object's weight, which acts away from the centre of the circle. PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D Consider the radial acceleration of the object at point B. rcos  

The radial forces acting on the object are the tension T in the rod, or string, which acts towards the centre of the circle, and the component mgcos  of the object's weight, which acts away from the centre of the circle. PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D Consider the radial acceleration of the object at point B. rcos  

The radial forces acting on the object are the tension T in the rod, or string, which acts towards the centre of the circle, and the component mgcos  of the object's weight, which acts away from the centre of the circle. PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D Consider the radial acceleration of the object at point B. rcos  

Since the object is executing circular motion with instantaneous tangential velocity v, it must experience an instantaneous acceleration towards the centre of the circle. Hence, using Newton's second law T – mgcos  = PE = 0 PE = -mgr PE = -mgrcos   T T u v A B mg D Consider the radial acceleration of the object at point B.  rcos 

Bob on a String – simple pendulum If the bob is given too greater velocity then it can swing above the suspension point and the string may/may not go slack. If it remains tight then it completes full circles, if it goes slack it becomes a projectile. See practical demo of bob on string

T mg  

Bob on a Rod – simple pendulum If the bob is given too greater velocity then it can swing above the suspension point but the rod means it cannot go slack. It completes full circles, or swings back and forth from side to side See practical demo of bob on rod

Bob inside a Cylinder If the bob is given too greater velocity then it can roll above the centre of the circle and the bob may/may not lose contact. If it remains in contact then it completes full circles, if it loses contact it becomes a projectile. See practical demo

Bob inside a Cylinder If the bob is given too greater velocity then it can roll above the centre of the circle and the bob may/may not lose contact. If it remains in contact then it completes full circles, if it loses contact it becomes a projectile. See practical demo

P.E. = –mgr K.E = ½ mu 2 P.E. = 0 K.E = ½ mv 1 2 P.E. = mgr K.E = ½ mv 2 2 Vertical Circles – bob on a wire. Two possibilities

P.E. = –mgr K.E = ½ mu 2 P.E. = 0 K.E = ½ mv 1 2 P.E. = mgrcos  K.E = 0 Vertical Circles – bob on a wire. Two possibilities