Showing posts with label Circular motion. Show all posts
Showing posts with label Circular motion. Show all posts
Wednesday, January 25, 2012
Wednesday, May 7, 2008
Concept Review - Chapter 7 Circular Motion
Angular position of the particle
If you visualize a particle moving in a radius of r with O as the centre of the circle. That means OP is equal to r. With O as origin mark OX on the X axis.
The position of P can be described by angle θ between Op and OX.
Angle θ is called the angular position of the particle
Angular velocity
The rate of change of angular position is called angular velocity.
ω = dθ/dt
Angular acceleration
The rate of change of angular velocity is angular acceleration.
Tangential accelartion
Radial unit vector
Tangential unit vector
Centripetal acceleration
Centripetal force
Centrifugal force
Colatitude
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
If you visualize a particle moving in a radius of r with O as the centre of the circle. That means OP is equal to r. With O as origin mark OX on the X axis.
The position of P can be described by angle θ between Op and OX.
Angle θ is called the angular position of the particle
Angular velocity
The rate of change of angular position is called angular velocity.
ω = dθ/dt
Angular acceleration
The rate of change of angular velocity is angular acceleration.
Tangential accelartion
Radial unit vector
Tangential unit vector
Centripetal acceleration
Centripetal force
Centrifugal force
Colatitude
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
Friday, February 8, 2008
IIT JEE Physics Formula Revision 7. Circular motion
Angular variables
θ = angular position of the particle
ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t
α = angular acceleration = d ω/dt = d²θ/dt²
If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:
θ = ω0t + ½ αt²
ω = ω0 + αt
ω² = ω0² + 2 α θ
Linear variables of circular motion
s = Linear distance traveled by the particle in circular motion
∆s = Linear distance traveled by the particle in circular motion in time ∆t
∆s = r∆θ
Where
r = radius of the circle over which the particle is moving
∆θ = angular displacement in time ∆t
∆s/∆t = r∆θ/∆t
v = r ω
where
v = linear speed of the particle
at = rate of change of speed of the particle in circular motion
at = dv/dt = rdω/dt = r α
Unit Vectors along the Radius and Tangent on a point on the circle on which the particle is moving.
If x-axis is horizontal and y axis is vertical (normal representation)
er = unit vector along the radius at a point on the circle
er = i cos θ+j sin θ
et = unit vector along the tangent at a point on the circle
et = - i sin θ+ j cos θ
Position vector of the point P
If the angular position is θ, x coordinate is r cos θ, and y coordinate is r sin θ.
Hence position vector is r = r(i cos θ + j sin θ)
Differentiating position vector with respect to time we get velocity
v = r ω(-i sin θ + j cos θ)
Differentiating v with respect to time
a = dv/dt = - ω²r er + dv/dt et
Uniform Circular Motion
If the particle moves in the circle with a uniform speed (v = constant), it is uniform
circular motion.
In this case dv/dt = 0
Hence
a = dv/dt = - ω²r er
Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is
a = ω²r = v²/r (as v = rω)
Nonuniform circular motion
In nonuniform circular motion, speed is not constant and the acceleration of the particle has both radial and tangential components.
a = - ω²r er + dv/dt et
the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt
The magnitude of the acceleration
a = √ (ar² + at²)
= √[( ω²r) ² + (dv/dt) ²]
The direction of this acceleration makes an angle α with the radius connecting the centre of the circle with the point at that instant.
Tan α = (dv/dt)/ (ω²r)
Forces in Circular Motion
If a particle of mass m is moving along a circle with uniform speed, the force acting on it has to be
F = ma = mv²/r = m ω²r
This force is called centripetal force.
Centrifugal Force
It is equal to the centripetal force
F = mv²/r = m ω²r
Circular Turnings on the Road
When a turn is made, the speed has to be less so that friction can make the vehicle turn in stead of skid
For a safe turn we need to have
v²/r = fs/M
Or M v²/r ≤ fs
As fs≤ μsMg
M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)
Instead of relying on friction, road are given an angle to provide centripetal force.
Tan θ = v²/gr
θ is the angle to be given to the road based v. Road are designed for assumed normal speed.
Effect of Earth’s Rotation on Apparent Weight
Axis of rotation of earth = The line joining the north and south pole = SN
Every point P on earth moves in circle.
Draw a perpendicular PC to SN from P.
Centre of the earth = O.
The distance between a point on the equator and centre of earth O = R radius of earth.
The angle between OP and PC is θ
θ is called the colatitude of the place where P is there.
If heavy particle of mass m is suspended through a string at P, the forces on it will be
1. gravitational attraction mg towards earth along PO
2. Centrifugal force = m ω² r
3. Tension in the string
The resultant of mg and m ω² r =
m√(g²+ ω4R²sin² θ-2g ω²R sin² θ) = mg’
The direction of this resultant force with OP is given by
Tan α = ω²Rsin θ cos θ/(g - ω²Rsin² θ)
A plumbline stays along g’
g’ = √(g²- ω²Rsin² θ(2g- ω²R ))
As 2g> ω²R; g’
At equator θ = 90° and
G’ = g- ω²R
At poles θ = 0° and hence
g’ = g
θ = angular position of the particle
ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t
α = angular acceleration = d ω/dt = d²θ/dt²
If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:
θ = ω0t + ½ αt²
ω = ω0 + αt
ω² = ω0² + 2 α θ
Linear variables of circular motion
s = Linear distance traveled by the particle in circular motion
∆s = Linear distance traveled by the particle in circular motion in time ∆t
∆s = r∆θ
Where
r = radius of the circle over which the particle is moving
∆θ = angular displacement in time ∆t
∆s/∆t = r∆θ/∆t
v = r ω
where
v = linear speed of the particle
at = rate of change of speed of the particle in circular motion
at = dv/dt = rdω/dt = r α
Unit Vectors along the Radius and Tangent on a point on the circle on which the particle is moving.
If x-axis is horizontal and y axis is vertical (normal representation)
er = unit vector along the radius at a point on the circle
er = i cos θ+j sin θ
et = unit vector along the tangent at a point on the circle
et = - i sin θ+ j cos θ
Position vector of the point P
If the angular position is θ, x coordinate is r cos θ, and y coordinate is r sin θ.
Hence position vector is r = r(i cos θ + j sin θ)
Differentiating position vector with respect to time we get velocity
v = r ω(-i sin θ + j cos θ)
Differentiating v with respect to time
a = dv/dt = - ω²r er + dv/dt et
Uniform Circular Motion
If the particle moves in the circle with a uniform speed (v = constant), it is uniform
circular motion.
In this case dv/dt = 0
Hence
a = dv/dt = - ω²r er
Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is
a = ω²r = v²/r (as v = rω)
Nonuniform circular motion
In nonuniform circular motion, speed is not constant and the acceleration of the particle has both radial and tangential components.
a = - ω²r er + dv/dt et
the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt
The magnitude of the acceleration
a = √ (ar² + at²)
= √[( ω²r) ² + (dv/dt) ²]
The direction of this acceleration makes an angle α with the radius connecting the centre of the circle with the point at that instant.
Tan α = (dv/dt)/ (ω²r)
Forces in Circular Motion
If a particle of mass m is moving along a circle with uniform speed, the force acting on it has to be
F = ma = mv²/r = m ω²r
This force is called centripetal force.
Centrifugal Force
It is equal to the centripetal force
F = mv²/r = m ω²r
Circular Turnings on the Road
When a turn is made, the speed has to be less so that friction can make the vehicle turn in stead of skid
For a safe turn we need to have
v²/r = fs/M
Or M v²/r ≤ fs
As fs≤ μsMg
M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)
Instead of relying on friction, road are given an angle to provide centripetal force.
Tan θ = v²/gr
θ is the angle to be given to the road based v. Road are designed for assumed normal speed.
Effect of Earth’s Rotation on Apparent Weight
Axis of rotation of earth = The line joining the north and south pole = SN
Every point P on earth moves in circle.
Draw a perpendicular PC to SN from P.
Centre of the earth = O.
The distance between a point on the equator and centre of earth O = R radius of earth.
The angle between OP and PC is θ
θ is called the colatitude of the place where P is there.
If heavy particle of mass m is suspended through a string at P, the forces on it will be
1. gravitational attraction mg towards earth along PO
2. Centrifugal force = m ω² r
3. Tension in the string
The resultant of mg and m ω² r =
m√(g²+ ω4R²sin² θ-2g ω²R sin² θ) = mg’
The direction of this resultant force with OP is given by
Tan α = ω²Rsin θ cos θ/(g - ω²Rsin² θ)
A plumbline stays along g’
g’ = √(g²- ω²Rsin² θ(2g- ω²R ))
As 2g> ω²R; g’
At equator θ = 90° and
G’ = g- ω²R
At poles θ = 0° and hence
g’ = g
Tuesday, October 23, 2007
Study guide H C Verma JEE Physics Ch. 7 CIRCULAR MOTION
JEE Syllabus
Uniform Circular motion;
-----------
Sections of the chapter
7.1 Angular variables
7.2 Unit vectors along the radius and the tangent
7.3 Acceleration in circular motion
7.4 Dynamics of circular motion
7.5 Circular turnings and banking of roads
7.6 Centrifugal force
7.7 Effect of earth's rotation on apprarent weight
------------
Study Plan
Day 1
7.1 Angular variables
Ex. 7.1, 7.2
7.2 Unit vectors along the radius and the tangent
Day 2
7.3 Acceleration in circular motion
Ex. 7.3,7.4
7.4 Dynamics of circular motion
Ex. 7.5
W.O.E. 3,4,5,7
Day 3
7.5 Circular turnings and banking of roads
Ex. 7.6
W.O.E. 1 and 2
Day 4
7.6 Centrifugal force
w.O.E. 7, 8,9,10,
Day 5
7.7 Effect of earth's rotation on apprarent weight
Ex. 7.7
w.O.E. 11 to 13
Day 6
Chapter 7
Objective I
Day 7
Chapter 7
Objective II
Day 8
Chapter 7
Exercises 1 to 5
Day 9
Chapter 7
Exercises 6 to 10
Day 10
Chapter 7
Exercises 11 to 15
Revision period
Chapter 7 Revision
Day 11
Chapter 7 Revision
Exercises: 16 to 18
Day 12
Chapter 7 Revision
Exercises: 19 to 21
Day 13
Chapter 7 Revision
Exercises: 22 to 24
Day 14
Chapter 7 Revision
Exercises: 25 to 27
Day 15
Chapter 7 Revision
Exercises: 28 to 30
Day 16
Chapter 7 Revision
Short answer questions: 1 to 10
Day 17
Chapter 7 Revision
Try some past JEE questions
Day 18
Chapter 7 Revision
Try some past JEE questions
Day 19
Chapter 7 Revision
Try some past JEE questions
Day 20
Chapter 7 Revision
Try some past JEE questions
Concepts covered
7.1 Angular variables
Angular position
angular velocity
Angular acceleration
Linear speed
Rate of change of speed
Tangential accelaration
7.2 Unit vectors along the radius and the tangent
radial unit vector
Tangential unit vector
7.3 Acceleration in circular motion
Uniform circular motion
Centripetal acceleration
radial component of acceleration
Tangential component of acceleration
7.4 Dynamics of circular motion
Centripetal force
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
---------
JEE Question 2007 Paper I
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2w using the entire potential energy of a spring compressed by a distance x-1 . Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x-2 . Both the discs rotate in the
clockwise direction
1. The ratio x-1 /x-2 is
(A) 2
(B)1/2
(C) SQRT(2)
(D)1/SQRT(2)
Answer C
2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t.The average frictional torque on one disc by the other during this period is
3.The loss of kinetic energy during the above process is
--------------------------
JEE question 2007 Paper II
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v^2/4g with respect to the initial position. The object is
(A) ring
(B) solid sphere
(C) hollow sphere
(D) Disc
Answer: D
-------------------------------
JEE Question Paper II 2007
Statement - 1
If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.
Because
Statement - 2
The linear momentum of an isolated system remains constant.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Right choice; D
S1 is incorrect. without external torque but with external force the centre can move.
S2 is correct.
----------------------
JEE 2006
A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be
(A) 2R/SQRT(15)
(B) R*(SQRT(2/15)
(C) 4R/SQRT(15)
(D) R/4
Answer: A
-------------------------
Uniform Circular motion;
-----------
Sections of the chapter
7.1 Angular variables
7.2 Unit vectors along the radius and the tangent
7.3 Acceleration in circular motion
7.4 Dynamics of circular motion
7.5 Circular turnings and banking of roads
7.6 Centrifugal force
7.7 Effect of earth's rotation on apprarent weight
------------
Study Plan
Day 1
7.1 Angular variables
Ex. 7.1, 7.2
7.2 Unit vectors along the radius and the tangent
Day 2
7.3 Acceleration in circular motion
Ex. 7.3,7.4
7.4 Dynamics of circular motion
Ex. 7.5
W.O.E. 3,4,5,7
Day 3
7.5 Circular turnings and banking of roads
Ex. 7.6
W.O.E. 1 and 2
Day 4
7.6 Centrifugal force
w.O.E. 7, 8,9,10,
Day 5
7.7 Effect of earth's rotation on apprarent weight
Ex. 7.7
w.O.E. 11 to 13
Day 6
Chapter 7
Objective I
Day 7
Chapter 7
Objective II
Day 8
Chapter 7
Exercises 1 to 5
Day 9
Chapter 7
Exercises 6 to 10
Day 10
Chapter 7
Exercises 11 to 15
Revision period
Chapter 7 Revision
Day 11
Chapter 7 Revision
Exercises: 16 to 18
Day 12
Chapter 7 Revision
Exercises: 19 to 21
Day 13
Chapter 7 Revision
Exercises: 22 to 24
Day 14
Chapter 7 Revision
Exercises: 25 to 27
Day 15
Chapter 7 Revision
Exercises: 28 to 30
Day 16
Chapter 7 Revision
Short answer questions: 1 to 10
Day 17
Chapter 7 Revision
Try some past JEE questions
Day 18
Chapter 7 Revision
Try some past JEE questions
Day 19
Chapter 7 Revision
Try some past JEE questions
Day 20
Chapter 7 Revision
Try some past JEE questions
Concepts covered
7.1 Angular variables
Angular position
angular velocity
Angular acceleration
Linear speed
Rate of change of speed
Tangential accelaration
7.2 Unit vectors along the radius and the tangent
radial unit vector
Tangential unit vector
7.3 Acceleration in circular motion
Uniform circular motion
Centripetal acceleration
radial component of acceleration
Tangential component of acceleration
7.4 Dynamics of circular motion
Centripetal force
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
---------
JEE Question 2007 Paper I
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2w using the entire potential energy of a spring compressed by a distance x-1 . Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x-2 . Both the discs rotate in the
clockwise direction
1. The ratio x-1 /x-2 is
(A) 2
(B)1/2
(C) SQRT(2)
(D)1/SQRT(2)
Answer C
2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t.The average frictional torque on one disc by the other during this period is
3.The loss of kinetic energy during the above process is
--------------------------
JEE question 2007 Paper II
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v^2/4g with respect to the initial position. The object is
(A) ring
(B) solid sphere
(C) hollow sphere
(D) Disc
Answer: D
-------------------------------
JEE Question Paper II 2007
Statement - 1
If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.
Because
Statement - 2
The linear momentum of an isolated system remains constant.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Right choice; D
S1 is incorrect. without external torque but with external force the centre can move.
S2 is correct.
----------------------
JEE 2006
A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be
(A) 2R/SQRT(15)
(B) R*(SQRT(2/15)
(C) 4R/SQRT(15)
(D) R/4
Answer: A
-------------------------
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