Slide 1 : believes in people who deliver the right solutions, and it proudly possesses them.
Slide 2 : www.aptacads.com
Slide 3 : In the session today Laws of Motion
Constraint relations
Non inertial frames (Pseudo force)
Circular motion & Centripetal force
Some interesting problems
Slide 4 : www.aptacads.com Let us start Constraint relation:
When we want to establish relation between
the accelerations of two or more objects in a
system using the length of the strings connecting
the objects.
Length of the string is a constraint because it is
not changing
Application: Commonly used in compound pulley
system
Slide 5 : www.aptacads.com Find the acceleration of the blocks in the figure, the pulley is mass less and string is light and inextensible. Example 1
Slide 6 : www.aptacads.com Method Identify a fixed point in the system.
Identify all the moving objects in the system.
Assign an instantaneous position to each moving object with respect to the fixed point.
Write the length of all the strings in terms of the instantaneous positions
Differentiate this relation twice to establish a relation between the acceleration of the objects
Slide 7 : www.aptacads.com Solution Let a = acceleration of mass m
And A = acceleration of block M and pulley Differentiating the equation twice w.r.t time, we get (Length of the string is constant) where L is the length of the string 2A – a = 0 or a = 2A (x2 – x1) + x2 = L or 2x2 – x1 = L
Slide 8 : www.aptacads.com Example - 2 In the arrangement shown in the figure, find the acceleration of the blocks . All the pulleys have negligible mass.
Slide 9 : www.aptacads.com Solution Choosing vertical downward direction as positive with origin at the center of pulley A Differentiating these equations twice w.r.t. time where L1 and L2 are the lengths of two strings Eliminating aB , we get
Slide 10 : www.aptacads.com Solution We have five unknowns namely a1,a2,a3,T1and T2 and five equations to solve them. The solution is left as an exercise for the students . The answer is
Slide 11 : www.aptacads.com Non inertial frame When the reference frame is an accelerated then
the frame becomes non inertial frame of reference.
Newton’s laws of motion are not valid in this frame.
Can we modify Newton laws so that they are valid in non-inertial frames also ? : Can we modify Newton laws so that they are valid in non-inertial frames also ? Yes, if we include the pseudo force in addition to all other real forces. The physically apparent but nonexistent force needed by an observer in a non inertial frame to make Newton's laws of motion hold true. What is a pseudo force ?
Slide 13 : Note:
Pseudo-force needs to be taken into account only when the problem is solved in a non-inertial frame.
There is no action-reaction pair for Pseudo-force.
Like gravitational force, pseudo-force is proportional to the mass of the body.
It acts in the direction opposite to the acceleration of the inertial frame.
Slide 14 : www.aptacads.com Example Figure shows a bob of mass m attached to the ceiling of an accelerated cart by means of a string. Find the value of angle that string makes with the vertical.
Slide 15 : www.aptacads.com Solution Analysis in the frame of the Cart (Non-inertial frame) Acceleration of bob (w.r.t cart) = 0
Slide 16 : www.aptacads.com Example Figure shows a block of mass m placed on a smooth wedge of mass M, which is free to slide along the horizontal surface. Block m is kept on the inclined surface (smooth) of the wedge. Find the acceleration of the wedge, so that the block does not slide on the wedge
Slide 17 : www.aptacads.com Solution
Slide 18 : www.aptacads.com General 2-D curvilinear motion Unlike i and j which refer to a fixed axis,
en and et do not have fixed directions. But an and at do have direct physical meaning. Normal component an arises from a change in the direction of velocity.
The tangential component at arises from a change in magnitude of velocity
Unit vectors along normal and tangential direction in circular motion : Unit vectors along normal and tangential direction in circular motion P N T
Slide 20 : www.aptacads.com Acceleration of a particle in circular motion:
Acceleration of a particle in circular motion: : Acceleration of a particle in circular motion: Acceleration is found by differentiating velocity w.r.t. time:
Slide 22 : www.aptacads.com Uniform Circular Motion In uniform circular motion, the particle moves with constant speed, thus Note:
Acceleration is directed towards the center .
Magnitude of acceleration is equal to v2/r This acceleration is also called centripetal acceleration. means seeking center
Non Uniform Circular Motion: : Non Uniform Circular Motion: Tangential direction: The magnitude of acceleration is
Note: : Note:
Slide 25 : www.aptacads.com Dynamics of uniform circular motion For a particle moving in a circle at constant speed,
A resultant non-zero force must provide the centripetal acceleration.
The forces responsible for this are called centripetal forces. 1. Choose one of the axis along the radius of the circle For solving problems: After resolving all the forces into components 2. And other axis perpendicular to the plane of the circle.
Uniform circular motion of a car on a horizontal surface: : Uniform circular motion of a car on a horizontal surface: R Which force provides the centripetal acceleration to a car moving on a circular track? Why do car skids on a slippery road when it attempts to take a turn ?
Slide 27 : Banking of Curves Why are roads banked at turns ?
Slide 28 : Example
Slide 29 : 1. n > w.
2. n = w.
3. n < w.
4. We can’t tell about n without knowing v. A car is rolling over the top of a hill at speed v. At this instant, Solution
Exercise : Exercise A particle of mass m slides down a fixed, frictionless sphere
of radius R, starting from rest at the top.
In terms of m, g, R. and ?, determine each of the
following for the particle while it is sliding on the sphere.
i. The kinetic energy of the particle
ii. The centripetal acceleration of the mass
iii. The tangential acceleration of the mass
b. Determine the value of ? at which the particle leaves the sphere.
Slide 31 : www.aptacads.com Thank you