ALL CIE PHYSICS DOUBTS HERE !!!

[elemis]

Please post all doubts here.

[ruby92]

Can someone please explain in detail how to derive the equation for centerpital acceleration?

[elemis]

Can someone please explain in detail how to derive the equation for centerpital acceleration?

http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml

http://www.youtube.com/watch?v=YRBRarbMCyE <-------- watch video first,

[ruby92]

thank you:)
I have two more doubts aswell atm.
from circular motion:
At an airshow an aircraft diving at the speed of 170ms-1 pulls out of the dive by moving in the arc of a circle at the bottom of the dive.
a) Calculate the minimum radius of this circle if the centerpital acceleration of the aircraft is not to exceed 5 times the acceleration of freefall.

From gravitaitonal fields:
The radius of the moons orbit is 3.84*10^8, and its period is 27.4 days.Use keplers law to calculate the period of the orbit of a satellite orbiting the earth just above the earths surface.

[S.M.A.T]

from circular motion:
At an airshow an aircraft diving at the speed of 170ms-1 pulls out of the dive by moving in the arc of a circle at the bottom of the dive.
a) Calculate the minimum radius of this circle if the centerpital acceleration of the aircraft is not to exceed 5 times the acceleration of freefall.

a=5g ms^-2
a=(v^2)/r
r=(v^2)/a
r=(170^2)/5g
r=(5780/g)m

[ruby92]

m/j 2008 paper 4 q1 b

[ashish]

m/j 2008 paper 4 q1 b

the centripetal force is provided by the frictional force..

note w=omega
W=weight

Fc= Frictional force
M(w^2)r =0.72W
M(w^2)(35/100) =0.72Mg
M cancel out

w^2=(0.72*9.81)/0.35

w=4.49 rad/s


frequency of rotation =  w/2pi
                             =0.7146
in 60 sec no. of rotation = 0.7146*60
=43

[ruby92]

THANK YOU
also could someone please help me with oct/nov 2009 paper 41 q1 bii

[ashish]

ok

Gravitational force = centripetal force
(GMm)/x^2 = Mw^2 x            (w=omega)
m is eliminated on both side
GM= w^2 x^3-----------1

we know that g is Gravitational field strength on earth surface

g=GM/R^2

BY making GM subject of formula

GM = gR^2------------

now
now replace GM by gR^2 in 1

gR^2 =w^2 x^3

[Ghost Of Highbury]

Question Attached.

Doubt : (i) (direction)

           2) For a force F of magnitude 150N, determine the force in S1.

I didn't understand the diagram very well. How exactly will the system move if a force is applied to the end of the lever. Also, how are equal forces produced in the strings? Could someone explain please?

[ashish]

Question Attached.

Doubt : (i) (direction)

           2) For a force F of magnitude 150N, determine the force in S1.

I didn't understand the diagram very well. How exactly will the system move if a force is applied to the end of the lever. Also, how are equal forces produced in the strings? Could someone explain please?

if the string would no have been there then the disc would have rotated upon itself!

just like opening nuts with a Cylinder Opener Key


2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)

moment = (30/100)* 150
=45Nm

couple is perpendicular distance  between the 2 forces * magnitude of one force

45=(12/100)*F
F=375N


sorry but i am getting problem while uploading pictures

the tension in the string would be toward the disc

[ashish]

thank you:)
I have two more doubts aswell atm.

From gravitaitonal fields:
The radius of the moons orbit is 3.84*10^8, and its period is 27.4 days.Use keplers law to calculate the period of the orbit of a satellite orbiting the earth just above the earths surface.

Keplers law T^2 is direcly proportional to R^3

therefore T^2/R^3 = constant

(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14

let the period be T

T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min

is that right?

[Ghost Of Highbury]

if the string would no have been there then the disc would have rotated upon itself!

just like opening nuts with a Cylinder Opener Key


2) the moment produce will be equal to the couple ( the two string will have opposite but equal forces)

moment = (30/100)* 150
=45Nm

couple is perpendicular distance  between the 2 forces * magnitude of one force

45=(12/100)*F
F=375N


sorry but i am getting problem while uploading pictures

the tension in the string would be toward the disc

k thanks a lot, + rep.

[ashish]

k thanks a lot, + rep.

thanks mate

[Deadly_king]

Keplers law T^2 is direcly proportional to R^3

therefore T^2/R^3 = constant

(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14

let the period be T

T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min

is that right?

Good job pal

keep it up

+rep