Revision of Horizontal Circular MotionThe angular velocity =

The tangential velocity =

Force towards the centre =

Time for 1 revolution =

w rads s-1

w is constant and does not change

v ms–1

v is constant and does not changemv2

r=mr ω2

2 πω

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 q measure the angular position of the object, measured from the downward vertical. Let u be the velocity of the object at q = 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.

aT will be negativei.e as q increases aT decreases

Formula for tangential accelerationSuppose the particle makes angle with the downward vertical after t seconds and has a tangential velocity vT and tangential acc aT. l = r (length of arc from core 2, where must be in radians) as r is a constant

For motion in a horizontal circle v = rwhere is the constant rate of change of

If varies as in vertical circles then = =

So vT = r where means

aT =

aT =

d d(rθ)= =

dt dt

l

d

dt

.

.

.

d

dt

...dv d(r )

rdt dt

..

r

dθ r

dt

vT

aT

vT = r

aT = r

ar =

vT = r

Variable VelocityConstant Velocity

aT = 0

ar = =

1) If = 2t5 , r = 0.3m find aT and ar after 2 seconds.

= 10t4

= 40t3

aT = r = 0.3×40t3 = 12t3 = 96ms-2 when t = 2.

also, ar = = 1602×0.3 = 7680ms-2 when t = 2.

.

..

.2 2 .

2v (r )( ) r

r r

..

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 = 0.2ar so ar = 6.51 ms-2

F = ma tangentially –0.2gsin30 = 0.2aT so aT = –4.9 ms-2

T=3

30o

30o

aT

0.2g

EX D1.1 pg 33 1–4

PE = 0

PE = -mgr

PE = -mgrcosq

qT

T

u

v

A

B

mg

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 rcosq. 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 –mgrcosq.

Consider an object moving around a vertical circle of radius r.

rcosq

q

Using conservation of energyEnergy at A = Energy at B

mu2 – mgr = mv2 – mgrcosq

which reduces to v2 = u2 + 2gr(cosq - 1)

PE = 0

PE = -mgr

PE = -mgrcosq

qT

T

u

v

A

B

mg

mg

D

rcosq

q

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 = -mgrcosq

qT

T

u

v

A

B

mg

mg

DConsider the radial acceleration of the object at point B.

rcosq

q

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 = -mgrcosq

qT

T

u

v

A

B

mg

mg

DConsider the radial acceleration of the object at point B.

rcosq

q

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 = -mgrcosq

qT

T

u

v

A

B

mg

mg

DConsider the radial acceleration of the object at point B.

rcosq

q

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 = -mgrcosq

qT

T

u

v

A

B

mg

mg

DConsider the radial acceleration of the object at point B.

2mvr

q

rcosq

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

q

q

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 CylinderIf 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 mu2

P.E. = 0 K.E = ½ mv12

P.E. = mgr K.E = ½ mv22

Vertical Circles – bob on a wire. Two possibilities

P.E. = –mgr K.E = ½ mu2

P.E. = 0 K.E = ½ mv12

P.E. = mgrcos K.E = 0

Vertical Circles – bob on a wire. Two possibilities