5.1 First law of motion (From HC Verma Part 1)
Study Plan
Day 1
5.1 First law of motion
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Points to be remembered
First law of motion
If the (vector) sum of all the forces acting on a particle is zero then and only then the particle remains unaccelerated (i.e., remains at rest or moves with constant velocity).
We can say in vector notation
a = 0 if and only if resultant force F = 0
A frame of reference in which Newton's first law is valid is called an inertial frame of reference.
A frame of reference in which Newton's first law is not valid is called a noninertial frame of reference. (Example: lamp in an elevator cabin whose cable had broken)
Example of lamp in an elevator whose cable had broken:
In the cabin when one measures with reference to the cabin, the lamp hanging from the ceiling has no acceleration. Hence the forces acting on the lamp, its weight (W) and the tension in the string supporting it are balancing each other. We can infer that W = T.
But for an observer on the ground, lamp is accelerating with acceleration g, when he considers the forces acting on the lamp as W and T once again, W is not equal to T as lamp is acceleating. Both cannot be right at the same time, and it means in one of the frames Newton's first law is not applicable.
Inertial frame: Hence inertial frame is a frame of reference in which Newton's first law is valid.
Is earth an inertial frame of reference?
Strictly it is not. But as a good approximation, earth can be taken as an inertial frame of reference.
All frames moving uniformly with respect to an inertial frame are themselves inertial.
This relation was derived from the relation that converts acceleration with respect to one frame into acceleration with respect to another frame. When the other frame is moving with uniform velocity it acceleration with respect to the frame in reference is zero.
Examples: A train moving with uniform velocity with respect to ground, a plane flying with uniform velocity with respect to a high etc. The sum of forces acting on a suit case kept on the shelves of them with turnout to be zero.
Concepts covered in the session
First law of motion
Frames of reference (concept from Chapter 3)
Inertial frame of reference
noninertial frame of reference
Inertial frames other than earth (frames moving with uniform velocity w.r.t earth)
Example 5.1 deals with forces acting on heavy particle hanging from a string fixed with the roof. the particle is stationery. The forces acting are pull of the earth downward due to gravity and pull of the string vertically upward.
Go through examples in worked out examples
1.
2.
Attempt questions in Objective questions (OBJ I)
1.
5.
Attempt questions in Objective questions (OBJ II)
1.
2.
3.
Sunday, August 31, 2008
Thursday, August 28, 2008
Angular Momentum - July Dec Revision
Principle of conservation of angular momentum
If the total external torque in a system is zero, its angular momentum remains constant.
Angular momentum of a particle about a point O is defined as
l = r × p
where
p = linear momentum
r = position vector of the particle from the given point O.
The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system.
If the total external torque in a system is zero, its angular momentum remains constant.
Angular momentum of a particle about a point O is defined as
l = r × p
where
p = linear momentum
r = position vector of the particle from the given point O.
The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system.
Wednesday, August 27, 2008
Kepler’s Laws of Planetary Motion - July Dec Revision
Kepler’s Laws of Planetary Motion
1. All planets move in elliptical orbits with the sun at a focus.
2. The radius vector from the sun to the planet sweeps equal area in equal time.
3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.
(Chapter: Gravitation)
1. All planets move in elliptical orbits with the sun at a focus.
Circular path is a special case of an ellipse when the major and minor axes are equal. For a circular apth, the planet should have velocity perpendicular to the line joining it with the sun and the magnitude has to be
v = √(GM/a)
If these conditions arenot satisfied, the planet moves in an ellipse.
2. The radius vector from the sun to the planet sweeps equal area in equal time.
For a circular orbit, as the speed of the particle remains constant, it will sweep equal area in equal time.
3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.
For circular orbits this law is proved.
Past JEE Questions
According to Kepler’s second law, the radius vector to a plant form the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of -----------------.
(JEE 1985)
Answer: Angular momentum
1. All planets move in elliptical orbits with the sun at a focus.
2. The radius vector from the sun to the planet sweeps equal area in equal time.
3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.
(Chapter: Gravitation)
1. All planets move in elliptical orbits with the sun at a focus.
Circular path is a special case of an ellipse when the major and minor axes are equal. For a circular apth, the planet should have velocity perpendicular to the line joining it with the sun and the magnitude has to be
v = √(GM/a)
If these conditions arenot satisfied, the planet moves in an ellipse.
2. The radius vector from the sun to the planet sweeps equal area in equal time.
For a circular orbit, as the speed of the particle remains constant, it will sweep equal area in equal time.
3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.
For circular orbits this law is proved.
Past JEE Questions
According to Kepler’s second law, the radius vector to a plant form the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of -----------------.
(JEE 1985)
Answer: Angular momentum
Tuesday, August 26, 2008
Collission - July-December 2008
Collision between two bodies - (For example two balls)
If no external force is acting on the system that is two ball, during whole process of colliosion (before the collision, during the collision, and after the collision) the momentum of the two-body system will remain constant.
If before the collision velocities are v1 and v2, masses m1 and m2
m1v1 + m2v2 = P
If during the collsion (at the instant) both have the same velocity V
(m1 + m2)*V = P
After the collision velocities are say v1' and v2'
m1v1' + m2v2' = P
In the case of energy of the system, assuming that there is no friction, the sum of the kinetic energy before the collision and after the collision will be same in case of elastic collision.
One example of elastic collision given in the book is a spring attached to the ball at the beginning which is getting hit from behind. due to the collision, the spring gets compressed and then expands.
Other example is balls perfectly elastic material. In the case of balls, there will be a deformation due to the collision and the surfaces will be in contact for some interval of time and travel together at the same speed. As the forces develop inside them the balls try to regain their original shape, and in the process push each other. The velocity of the front ball increases while that of the rear ball decreases and the balls separate. After the separation, the balls regain their original shape so that the elastic potential energy is completely converted back into kinetic energy.
Thus during an elastic collision, the initial kinetic energy of the two body system is equal to the final kinetic energy, but it does not remain constant as during the collision some kinetic energy is converted into elastic potential energy.
In the case of inelastic collsion, the two deformed balls have no tendency to regain the shape and tend to remain in contact after the deformation and move with same speed. The kinetic energy of the two body system decreases at the time of deformation and remains constant thereafter.
Collisions – Psat JEE Problems
1. A ball hits the floor and rebounds after an inelastic collision. In this case
a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.
(More than one alternative may be correct)
JEE 1986
Answer ( c ) and (d)
Note: total kinetic energy is not conserved but total energy is conserved.
2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .
(JEE 1987)
answer (3/2)mv²
Solution;
As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum
2mv cos 45° = 2mv(1/√2) = √2 mv
if the velocity of fragment with 2m mass is V
then 2 mV =/√2mv
V = v/√2
Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²
= (3/2)mv²
If no external force is acting on the system that is two ball, during whole process of colliosion (before the collision, during the collision, and after the collision) the momentum of the two-body system will remain constant.
If before the collision velocities are v1 and v2, masses m1 and m2
m1v1 + m2v2 = P
If during the collsion (at the instant) both have the same velocity V
(m1 + m2)*V = P
After the collision velocities are say v1' and v2'
m1v1' + m2v2' = P
In the case of energy of the system, assuming that there is no friction, the sum of the kinetic energy before the collision and after the collision will be same in case of elastic collision.
One example of elastic collision given in the book is a spring attached to the ball at the beginning which is getting hit from behind. due to the collision, the spring gets compressed and then expands.
Other example is balls perfectly elastic material. In the case of balls, there will be a deformation due to the collision and the surfaces will be in contact for some interval of time and travel together at the same speed. As the forces develop inside them the balls try to regain their original shape, and in the process push each other. The velocity of the front ball increases while that of the rear ball decreases and the balls separate. After the separation, the balls regain their original shape so that the elastic potential energy is completely converted back into kinetic energy.
Thus during an elastic collision, the initial kinetic energy of the two body system is equal to the final kinetic energy, but it does not remain constant as during the collision some kinetic energy is converted into elastic potential energy.
In the case of inelastic collsion, the two deformed balls have no tendency to regain the shape and tend to remain in contact after the deformation and move with same speed. The kinetic energy of the two body system decreases at the time of deformation and remains constant thereafter.
Collisions – Psat JEE Problems
1. A ball hits the floor and rebounds after an inelastic collision. In this case
a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.
(More than one alternative may be correct)
JEE 1986
Answer ( c ) and (d)
Note: total kinetic energy is not conserved but total energy is conserved.
2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .
(JEE 1987)
answer (3/2)mv²
Solution;
As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum
2mv cos 45° = 2mv(1/√2) = √2 mv
if the velocity of fragment with 2m mass is V
then 2 mV =/√2mv
V = v/√2
Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²
= (3/2)mv²
Saturday, August 23, 2008
Moment of Inertia - July -Dec 2008 Revision
Moment of Inertia
Torque created by external forces in rotating motion
When a particle is rotating it has tangential acceleration and radial acceleration.
Radial acceleration = ω²r
Hence radial force acting on it = m ω²r
Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα
Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α
In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles
Total torque on the body = Γ(total) = Σ miri²α
Σ miri² is called as moment of inertia.
Moment of inertia can be calculated using the above formula for collection of discrete particles.
If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.
We find ∫ r²dm under proper limits to get the moment of inertia of the body.
r²dm is the moment of inertia of the small element.
Determination of moment of inertia of representative bodies.
1. Uniform rod about a perpendicular
M = total mass of the body
l = length of the body
I = Ml²/12
I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)
dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12
2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.
M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b
I = Ml²/12
If the axis of line is parallel to l
I = Mb²/12
3. Circular ring
M = total mass of the body
R = radius
I = MR²
4. Uniform circular plate
M = total mass of the body
R = radius
I = MR²/2
5. Hollow cylinder about its axis
M = total mass of the body
R = radius
I = MR²
6. Uniform solid cylinder about its axis.
M = total mass of the body
R = radius
I = MR²/2
7. Hollow sphere about a diameter
M = total mass of the body
R = radius
I =(2/3) MR²
8. Uniform solid sphere about a diameter
M = total mass of the body
R = radius
I = (2/5) MR²
Parallel axis theorem
Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes).
Past JEE problem
1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is
a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²
JEE 2005
Answer (a)
Solution
Mass of the disc removed
Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²
Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M
Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²
According to the parallel axis theorem
Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²
Moment of inertia of bigger disc before cutting the piece
= ½ (9M)(r²)
Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²
Torque created by external forces in rotating motion
When a particle is rotating it has tangential acceleration and radial acceleration.
Radial acceleration = ω²r
Hence radial force acting on it = m ω²r
Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα
Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α
In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles
Total torque on the body = Γ(total) = Σ miri²α
Σ miri² is called as moment of inertia.
Moment of inertia can be calculated using the above formula for collection of discrete particles.
If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.
We find ∫ r²dm under proper limits to get the moment of inertia of the body.
r²dm is the moment of inertia of the small element.
Determination of moment of inertia of representative bodies.
1. Uniform rod about a perpendicular
M = total mass of the body
l = length of the body
I = Ml²/12
I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)
dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12
2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.
M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b
I = Ml²/12
If the axis of line is parallel to l
I = Mb²/12
3. Circular ring
M = total mass of the body
R = radius
I = MR²
4. Uniform circular plate
M = total mass of the body
R = radius
I = MR²/2
5. Hollow cylinder about its axis
M = total mass of the body
R = radius
I = MR²
6. Uniform solid cylinder about its axis.
M = total mass of the body
R = radius
I = MR²/2
7. Hollow sphere about a diameter
M = total mass of the body
R = radius
I =(2/3) MR²
8. Uniform solid sphere about a diameter
M = total mass of the body
R = radius
I = (2/5) MR²
Parallel axis theorem
Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes).
Past JEE problem
1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is
a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²
JEE 2005
Answer (a)
Solution
Mass of the disc removed
Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²
Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M
Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²
According to the parallel axis theorem
Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²
Moment of inertia of bigger disc before cutting the piece
= ½ (9M)(r²)
Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²
Saturday, August 9, 2008
Physics Formula Revision Rotational mechanics July - Dec 2008
Formula Sheet – rotational mechanics
Rotational kinematics
Angular variables
θ = angular position of the particle
ω = angular velocity = dθ/dt = lim∆t→0 ∆θ/∆t
α = angular acceleration = dω/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
where
ω0 = angular velocity at the beginning
Relation between the linear motion of a particle of a rigid body and rotation of the rigid body
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 (These relations are from the chapter of circular motion)
Rotational kinematics
Angular variables
θ = angular position of the particle
ω = angular velocity = dθ/dt = lim∆t→0 ∆θ/∆t
α = angular acceleration = dω/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
where
ω0 = angular velocity at the beginning
Relation between the linear motion of a particle of a rigid body and rotation of the rigid body
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 (These relations are from the chapter of circular motion)
Rotational Mechanics - July - Dec Revision 2008
Rotation of a rigid body
If each particle of a rigid body moves in a circle, with centres of all the circles on a straight line and with planes of the circles perpendicular to this line we say that the body is rotating about this line. The straight line is called the axis of rotation. (Particle makes circular motion. Rigid body makes rotation.)
Kinematics of rotation of rigid body
For a rigid body let the axis of rotation be Z-axis.
At time t =0, let particle P be at P0.
Perpendicular to axis of rotation from P0 be PQ. (Q is on the axis)
If at time t Particle P moves P1 and angle P0Q P1 = θ.
Hence the particle has rotated through θ.
All particles have rotated through θ.
We can say the whole rigid body has rotated through angle θ.
The angular position of the body at time t is θ.
If P has made a complete revolution its circular path, every particle will do so and hence rigid body has done so. We can say rigid body made a complete revolution and it has rotated through an angle of 2 π radians.
Hence, rotation of a rigid body is measured by the rotation of a line PQ (P is a particle on the rigid body and Q is a point on the axis of rotation and PQ is perpendicular from P to the axis of rotation).
As the rotation of rigid body is defined in terms of the circular motion of a particle on the rigid body, kinematics of circular motion of particle becomes applicable to rotation.
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
Where
ω0 is velocity at time t = 0.
Given the axis of rotation, the body can rotate in two directions – clockwise or anticlockwise. One of the directions has to be defined as positive direction according to the convenience of the problem.
The SI unit for angular velocity is radian/sec (rad/s).
One revolution/sec = 2 π radian/sec
Similar to the circular motion of the particle, in rotation for a particle P
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
Rotational dynamics
In rotation of a body the resultant force due to external forces is zero, but the resultant of
Torque produced by the external forces is nonzero and this torque produces rotation motion.
Torque of a force about the axis of rotation
First we define torque a force about a point.
For a force F acting on a particle P, to find torque about a point O, define the position vector of P with respect to O. Let this position vector be r.
Torque of force F about O = Γ = F × r.
This is vector product of two vectors hence a vector quantity, as per the rules of vector product, the direction of Γ will be perpendicular to to F and r.
When the torque about an axis of rotation is to be determined, select a point on the axis of the rotation and find the torque of the force acting on a particle about this point. Find the angle between the axis and the line joining the point on axis (about which the torque is calculated) and the particle (one which the force is acting). Let the angle be θ.
Torque about the axis due to a force is the component along the axis, of the torque of the force about a point on the axis.
Magnitude of the torque = | F × r | cos θ
The torque about the axis is same, even if different points are chosen along the axis for determining the torque of the force about those points.
Some special cases of relation between force and the axis of rotation.
1. Force is parallel to the axis of rotation.
Torque along the axis is zero.
2. F and r are collinear. The torque about O is zero and the torque about axis is zero.
3. Force and axis are perpendicular but they do not intersect. In three dimensions, two lines may be perpendicular without intersecting. Example: A vertical line on
a wall and a horizontal line on the opposite wall.
In this case torque about the axis is equal to Force multiplied by the perpendicular to axis from the force direction (line along which the force is acting).
Torque produced by forces on a particle
A particle having circular motion will have two forces acting on it.
One force produces tangential acceleration dv/dt in it. Hence the force named tangential force is 'ma' = mdv/dt = mrα
This force creates a torque of mr²α
the other force creates radial acceleration or centripetal acceleration ω²r. Hence the force named radial force is mω²r
As intersects the axis of rotation, the torque produced by it is zero.
Hence total torque produced by a rigid body consisting of n particles is
Г(total) = Σmiri²α
= αΣmiri² as α is same for all particles
Let I = Σmiri²
Г(total) = Iα
Quantity I is called moment of inertia of the body about the axis of rotation.
I = Σmiri²
where
mi = mass of the ith particle
ri = perpendicular distance of ith particle from the axis of rotation.
Angular momentum
Conservation of angular momentum
Work done by a torque
Power delivered by a torque
Moment of inertia
Moment of Inertia
Torque created by external forces in rotating motion
When a particle is rotating it has tangential acceleration and radial acceleration.
Radial acceleration = ω²r
Hence radial force acting on it = m ω²r
Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα
Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α
In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles
Total torque on the body = Γ(total) = Σ miri²α
Σ miri² is called as moment of inertia.
Moment of inertia can be calculated using the above formula for collection of discrete particles.
If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.
We find ∫ r²dm under proper limits to get the moment of inertia of the body.
r²dm is the moment of inertia of the small element.
Determination of moment of inertia of representative bodies.
1. Uniform rod about a perpendicular
M = total mass of the body
l = length of the body
I = Ml²/12
I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)
dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12
2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.
M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b
I = Ml²/12
If the axis of line is parallel to l
I = Mb²/12
3. Circular ring
M = total mass of the body
R = radius
I = MR²
4. Uniform circular plate
M = total mass of the body
R = radius
I = MR²/2
5. Hollow cylinder about its axis
M = total mass of the body
R = radius
I = MR²
6. Uniform solid cylinder about its axis.
M = total mass of the body
R = radius
I = MR²/2
7. Hollow sphere about a diameter
M = total mass of the body
R = radius
I =(2/3) MR²
8. Uniform solid sphere about a diameter
M = total mass of the body
R = radius
I = (2/5) MR²
Moment of inertia theorems
Theorem of parallel axes
Thoerem of perpendicular axes
Combined rotation and translation
Rolling
If each particle of a rigid body moves in a circle, with centres of all the circles on a straight line and with planes of the circles perpendicular to this line we say that the body is rotating about this line. The straight line is called the axis of rotation. (Particle makes circular motion. Rigid body makes rotation.)
Kinematics of rotation of rigid body
For a rigid body let the axis of rotation be Z-axis.
At time t =0, let particle P be at P0.
Perpendicular to axis of rotation from P0 be PQ. (Q is on the axis)
If at time t Particle P moves P1 and angle P0Q P1 = θ.
Hence the particle has rotated through θ.
All particles have rotated through θ.
We can say the whole rigid body has rotated through angle θ.
The angular position of the body at time t is θ.
If P has made a complete revolution its circular path, every particle will do so and hence rigid body has done so. We can say rigid body made a complete revolution and it has rotated through an angle of 2 π radians.
Hence, rotation of a rigid body is measured by the rotation of a line PQ (P is a particle on the rigid body and Q is a point on the axis of rotation and PQ is perpendicular from P to the axis of rotation).
As the rotation of rigid body is defined in terms of the circular motion of a particle on the rigid body, kinematics of circular motion of particle becomes applicable to rotation.
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
Where
ω0 is velocity at time t = 0.
Given the axis of rotation, the body can rotate in two directions – clockwise or anticlockwise. One of the directions has to be defined as positive direction according to the convenience of the problem.
The SI unit for angular velocity is radian/sec (rad/s).
One revolution/sec = 2 π radian/sec
Similar to the circular motion of the particle, in rotation for a particle P
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
Rotational dynamics
In rotation of a body the resultant force due to external forces is zero, but the resultant of
Torque produced by the external forces is nonzero and this torque produces rotation motion.
Torque of a force about the axis of rotation
First we define torque a force about a point.
For a force F acting on a particle P, to find torque about a point O, define the position vector of P with respect to O. Let this position vector be r.
Torque of force F about O = Γ = F × r.
This is vector product of two vectors hence a vector quantity, as per the rules of vector product, the direction of Γ will be perpendicular to to F and r.
When the torque about an axis of rotation is to be determined, select a point on the axis of the rotation and find the torque of the force acting on a particle about this point. Find the angle between the axis and the line joining the point on axis (about which the torque is calculated) and the particle (one which the force is acting). Let the angle be θ.
Torque about the axis due to a force is the component along the axis, of the torque of the force about a point on the axis.
Magnitude of the torque = | F × r | cos θ
The torque about the axis is same, even if different points are chosen along the axis for determining the torque of the force about those points.
Some special cases of relation between force and the axis of rotation.
1. Force is parallel to the axis of rotation.
Torque along the axis is zero.
2. F and r are collinear. The torque about O is zero and the torque about axis is zero.
3. Force and axis are perpendicular but they do not intersect. In three dimensions, two lines may be perpendicular without intersecting. Example: A vertical line on
a wall and a horizontal line on the opposite wall.
In this case torque about the axis is equal to Force multiplied by the perpendicular to axis from the force direction (line along which the force is acting).
Torque produced by forces on a particle
A particle having circular motion will have two forces acting on it.
One force produces tangential acceleration dv/dt in it. Hence the force named tangential force is 'ma' = mdv/dt = mrα
This force creates a torque of mr²α
the other force creates radial acceleration or centripetal acceleration ω²r. Hence the force named radial force is mω²r
As intersects the axis of rotation, the torque produced by it is zero.
Hence total torque produced by a rigid body consisting of n particles is
Г(total) = Σmiri²α
= αΣmiri² as α is same for all particles
Let I = Σmiri²
Г(total) = Iα
Quantity I is called moment of inertia of the body about the axis of rotation.
I = Σmiri²
where
mi = mass of the ith particle
ri = perpendicular distance of ith particle from the axis of rotation.
Angular momentum
Conservation of angular momentum
Work done by a torque
Power delivered by a torque
Moment of inertia
Moment of Inertia
Torque created by external forces in rotating motion
When a particle is rotating it has tangential acceleration and radial acceleration.
Radial acceleration = ω²r
Hence radial force acting on it = m ω²r
Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα
Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α
In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles
Total torque on the body = Γ(total) = Σ miri²α
Σ miri² is called as moment of inertia.
Moment of inertia can be calculated using the above formula for collection of discrete particles.
If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.
We find ∫ r²dm under proper limits to get the moment of inertia of the body.
r²dm is the moment of inertia of the small element.
Determination of moment of inertia of representative bodies.
1. Uniform rod about a perpendicular
M = total mass of the body
l = length of the body
I = Ml²/12
I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)
dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12
2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.
M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b
I = Ml²/12
If the axis of line is parallel to l
I = Mb²/12
3. Circular ring
M = total mass of the body
R = radius
I = MR²
4. Uniform circular plate
M = total mass of the body
R = radius
I = MR²/2
5. Hollow cylinder about its axis
M = total mass of the body
R = radius
I = MR²
6. Uniform solid cylinder about its axis.
M = total mass of the body
R = radius
I = MR²/2
7. Hollow sphere about a diameter
M = total mass of the body
R = radius
I =(2/3) MR²
8. Uniform solid sphere about a diameter
M = total mass of the body
R = radius
I = (2/5) MR²
Moment of inertia theorems
Theorem of parallel axes
Thoerem of perpendicular axes
Combined rotation and translation
Rolling
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