Sunday, January 31, 2016
JEE - CBSE - Physics Ch. 1 INTRODUCTION TO PHYSICS - Revision Questions
This blog was started along with my blog. Still being updated.
http://physicsgoeasy.blogspot.in/
Test on all chapters
http://www.mtel.nesinc.com/pdfs/ma_fld011_practice_test.pdf
http://www.topperlearning.com/study/cbse/class-11/physics/multiple-choice-questions/units-and-measurement/b101c3s4e23ch121#begin-test
Sunday, January 10, 2016
Concept Review - Chapter 9 Centre of Mass, Collision
Centre of mass
Definition
Centre of mass of several groups of particles
Centre of mass of continuous bodies
Motion of centre of mass
Linear momentum
Principle of conservation of linear momentum
Collision
Elastic collision
Inelastic collision
Coefficient of restitution
Impulse
Impulse force
Centre of mass - Linear momentum - Videos
Updated 10 Jan 2016, 7 May 2008
Definition
Centre of mass of several groups of particles
Centre of mass of continuous bodies
Motion of centre of mass
Linear momentum
Principle of conservation of linear momentum
Collision
Elastic collision
Inelastic collision
Coefficient of restitution
Impulse
Impulse force
Centre of mass - Linear momentum - Videos
Updated 10 Jan 2016, 7 May 2008
Concept Review - Chapter 10 Rotational Mechanics
Rotation of a rigid body
Kinematics of rotation of rigid body
Torque of a force about 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 theorems
Theorem of parallel axes
Theorem 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
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)
Updated 10 Jan 2016, 7 May 2008
Kinematics of rotation of rigid body
Torque of a force about 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 theorems
Theorem of parallel axes
Theorem of perpendicular axes
Combined rotation and translation
Rolling
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
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)
Updated 10 Jan 2016, 7 May 2008
Saturday, January 9, 2016
Concept Review - Chapter 8 Work and Energy
________________
________________
________________
________________
Physics Pathasala upload
________________
________________
________________
________________
Etoos
Updated 9 Jan 2016, 7 May 2008
Subscribe to:
Posts (Atom)