/ 780. 360 - = 2T, Centripetal force Vertical circular motion
degrees e Radians 180 - = 1 • always directed towards centre In circular motion the speed mass on sprinG 1 - position
t
✗ 180. 90 - = • perpendicular to velocity of an object remains the same. mg
R ma = mg + R
2
F = mV² a =Y Because: net force is perpendicular 024 mV? = mgt R R = mV? -mg
↑R
T- J
r to direction of motion (velocity) 2 2- "position 2
mg
R
V-2TR In circular motion, This means that no work is done by my' = R-mg
f = l linear velocity changes the force so no change in speed. forces in opposite direction
linear speed is constant W- f ✗ cos@ cos 90' = Of R- my' + mg
We DO W-2¥ or If as velocity is a vector so direction W=fxO = 0 perpendicular
At changes. The centripetal acceleration v=wr
V-Wr changes the direction. F=M,V² = m (Wr)? = mw²r² = mw²r Circular motion at an ANGLE
r r R 0=0
RCOSO o
F
EXPERIMENT: • Keep R constant a =L a-w²r R(sino) vertical: Rcoso - weight - mg
vary mass r A horizontal: Rsino = my ² = centripetal force
string 2
- Measure m with a mass balance HOW TO DETERMINE T/F: spring Pendulum
R
iinder calculate weight which acts as We use a fiducial marker Toso o
Tcoso = mg T-%:O
centripetal force using F= mg to mark the point where Tsino = my- :. If sino =#r²
Isino
- measure radius with a ruler the spring and mass is in fiducial
miss r marker
- Time at least 10 time periods with equilibrium. Every time 9 ¾:-÷
• Plot : a stop clock to reduce percentage uncertainty. mass passes the mark
graph f- against v2 - Divide by 10 for 1 time period. that = 1 time period. gtano =#→ ✓ = grtano
draw line of best fit - calculate V = 2#R • displace the mass
if centripetal force is true • measure at least 10 Time periods from equilibrium position. SHM GRAPHS:
a straight line through origin - calculate v2 for different values of mass. • Take all measurements from eye level to avoid PARALLAX ERROR
x
occurs • divide measurement by 10 to get time period. grad = DX =
F = MV² to y= F • Take multiple measurements and then find mean to improve Δt
r x = V2 accuracy. t
y = moc + C grad = mass • calculate frequency 8=7/ T
radius
angular frequency: W=2Tf
m-gradient xr plot the gradient
SOLUTIONS TO SHM: of x-t graph
Terms: a = -wax on the V-t graph
Always Put
displacement (x): distance from oscillator to equilibrium in • gradient is decreasing
use when,t Acos (wt)
amplitude (A): Max displacement to equilibrium rad:{sulator x-t graph so we must
time period (T): time it takes to complete 7 full oscillation. or oscillator at amplitude start curve in negativ
frequency (f): number of oscillations per unit time a
Angular frequency (w): rate of change of angular displacement. x = A sin (wt) plot gradie
a- 1¥
w=§♀ = = 278 oscillator at equilibrium of V-t gr
t
E. g
SIMPLE HARMONIC MOTION: 7 a α oc a spring oscillates with f=5Hz and A- 4cm. The
a = -w²x 2 a is directed towards equilibrium position spring is Initially released from its amplitude. Find Free Oscillations Forced Oscillati
a a =-W² X to the displacement 2 seconds after release.
G y-ma← ,
+ c steady decrease x-Acos (wt) W=2TIf x-Acos (Wt)
degrees e Radians 180 - = 1 • always directed towards centre In circular motion the speed mass on sprinG 1 - position
t
✗ 180. 90 - = • perpendicular to velocity of an object remains the same. mg
R ma = mg + R
2
F = mV² a =Y Because: net force is perpendicular 024 mV? = mgt R R = mV? -mg
↑R
T- J
r to direction of motion (velocity) 2 2- "position 2
mg
R
V-2TR In circular motion, This means that no work is done by my' = R-mg
f = l linear velocity changes the force so no change in speed. forces in opposite direction
linear speed is constant W- f ✗ cos@ cos 90' = Of R- my' + mg
We DO W-2¥ or If as velocity is a vector so direction W=fxO = 0 perpendicular
At changes. The centripetal acceleration v=wr
V-Wr changes the direction. F=M,V² = m (Wr)? = mw²r² = mw²r Circular motion at an ANGLE
r r R 0=0
RCOSO o
F
EXPERIMENT: • Keep R constant a =L a-w²r R(sino) vertical: Rcoso - weight - mg
vary mass r A horizontal: Rsino = my ² = centripetal force
string 2
- Measure m with a mass balance HOW TO DETERMINE T/F: spring Pendulum
R
iinder calculate weight which acts as We use a fiducial marker Toso o
Tcoso = mg T-%:O
centripetal force using F= mg to mark the point where Tsino = my- :. If sino =#r²
Isino
- measure radius with a ruler the spring and mass is in fiducial
miss r marker
- Time at least 10 time periods with equilibrium. Every time 9 ¾:-÷
• Plot : a stop clock to reduce percentage uncertainty. mass passes the mark
graph f- against v2 - Divide by 10 for 1 time period. that = 1 time period. gtano =#→ ✓ = grtano
draw line of best fit - calculate V = 2#R • displace the mass
if centripetal force is true • measure at least 10 Time periods from equilibrium position. SHM GRAPHS:
a straight line through origin - calculate v2 for different values of mass. • Take all measurements from eye level to avoid PARALLAX ERROR
x
occurs • divide measurement by 10 to get time period. grad = DX =
F = MV² to y= F • Take multiple measurements and then find mean to improve Δt
r x = V2 accuracy. t
y = moc + C grad = mass • calculate frequency 8=7/ T
radius
angular frequency: W=2Tf
m-gradient xr plot the gradient
SOLUTIONS TO SHM: of x-t graph
Terms: a = -wax on the V-t graph
Always Put
displacement (x): distance from oscillator to equilibrium in • gradient is decreasing
use when,t Acos (wt)
amplitude (A): Max displacement to equilibrium rad:{sulator x-t graph so we must
time period (T): time it takes to complete 7 full oscillation. or oscillator at amplitude start curve in negativ
frequency (f): number of oscillations per unit time a
Angular frequency (w): rate of change of angular displacement. x = A sin (wt) plot gradie
a- 1¥
w=§♀ = = 278 oscillator at equilibrium of V-t gr
t
E. g
SIMPLE HARMONIC MOTION: 7 a α oc a spring oscillates with f=5Hz and A- 4cm. The
a = -w²x 2 a is directed towards equilibrium position spring is Initially released from its amplitude. Find Free Oscillations Forced Oscillati
a a =-W² X to the displacement 2 seconds after release.
G y-ma← ,
+ c steady decrease x-Acos (wt) W=2TIf x-Acos (Wt)
