Uniform circular motion

Rolling resistance, sometimes called rolling friction, is the resistance that occurs when an object (e.g a wheel or tire) rolls. It is much smaller than sliding friction except for special cases like ice skating. It is caused by the deformation of the wheel or tire or the deformation of the ground. It depends very much on the material of the wheel or tire and the sort of ground. For example, rubber will give a bigger rolling friction than steel. Also, sand on the ground will give more rolling friction than concrete. A vehicle rolling will gradually slow down due to rolling friction, but a train with steel wheels running on steel rails will roll much further than a car or truck with rubber tires running on pavement, even when differences in mass and momentum are accounted for.

Factors contributing towards rolling friction

Several factors affect the magnitude of rolling friction a tire generates:

Material - Tires with higher sulfur content tend to have a lower rolling friction. This is one strategy that most hybrid car vendors use to improve fuel efficiency.
Dimensions - rolling friction is related to the flex of sidewalls and the contact area of the tire.
Extent of inflation - Lower pressure in tires results in more flexing of sidewalls and higher rolling friction.
Over inflating tires (such a bicycle tires) may not lower the overall rolling resistance as the tire may skip and hop over the road surface. Traction is sacrificed, and overall rolling friction may not be reduced as the wheel rotational speed changes and slippage increases.
Tread thickness has much to do with rolling resistance. The thicker the tread, the higher the rolling resistance. Thus, the "fastest" bicycle tires have very little tread and heavy duty trucks get the best fuel economy as the tire tread wears out.
Smaller wheels, all else being equal, have higher rolling resistance than larger wheels

If mass of moon is already given (point 2) in the option then why do u bother to find it again? Please check your question again.

I think in step 2 it should be mass of earth rather than that of moon.

So if this is the case then to find mass of moon you can use Kepler's Law of planetary motion

T² = 4

²R³/GM

Relation between linear velocity and Angular velocity of a body about any point is given by

= (r x v) / I r I²

where v is linear velocity and r is position vector of the body w.r.t point about which we intend to find angular velocity

Further if the point is on the line of a velocity vector then the angle

is zero so sin

= 0, implying that angular velocity vanishes so

= 0

How can a bullet weigh 20kg?

Just take it on a light note. I know its a typing mistake. Anyway i consider the mass to be 20g for further calculations

Mass of bullet = m₁= 20g = 20/1000 kg = 0.02kg

Mass of block = m₂

Applying law of conservation of momentum after the bullet emerges with velocity 100m/s we get

(0.02)*500 = m₂v + (0.02)*100

or v = 8/m₂

Now block travels a distance of 20cm = 0.2m before coming to rest

using, v²- u²= 2as (put u = 8/m₂ and v = 0)

Now, Find accelertion(-ve) offered by frictional force(F =

mg) . Use frictional force equation

to find the coefficient of friction.

By two body oscillators if you mean coupled oscillations. Then examples are

1) two masses at the two ends of a spring and thus coupled by means of spring

2) In a lattice or solid the neighboring atoms are coupled to each other and thus the motion of one influences the motion of other neighboring atoms.

3) Similarly we talk about coupled harmonic oscillator.

Moreover Coupled Oscillations occur when two or more oscillating systems are connected in such a manner as to allow motion energy to be exchanged between them. Coupled oscillators occur in nature (e.g., the moon and earth orbiting each other) or can be found in man-made devices (such as with the pacemaker).

Coupled oscillations

The harmonic oscillator, and the more complicated systems for which it stands as a simple model, has a single degree of freedom. More complicated systems have more degrees of freedom, for example two masses and two springs. In such cases, energy is converted between the respective inertias of each degree of freedom and the several restoring forces in the system. This leads to a coupling of the oscillations of the individual degrees of freedom. For example, two pendulum clocks mounted on a common wall will tend to synchronise. The apparent motions of the individual oscillations typically appears very

Spring-coupled masses

Consider the two degree of freedom dynamical system pictured in folowing Figure. In this system, two point objects of mass

are free to move in one dimension. Furthermore, the masses are connected together by a spring of spring constant

, and are also each attached to fixed supports via springs of spring constant

.

**Spring Coupled masses**

Let

and

be the displacements of the first and second masses, respectively, from the equilibrium state. If follows that the extensions of the left-hand, middle, and right-hand springs are

,

, and

, respectively. The kinetic energy of the system takes the form

$egin{displaymath} K = rac{m}{2} ,(dot{q}_1^{,2} + dot{q}_2^{,2}), end{displaymath}$

whereas the potential energy is written

$egin{displaymath} U= rac{1}{2}left[k',q_1^{,2} + k,(q_2-q_1)^2+ k',q_2^{,2} ight]. end{displaymath}$

The above expression can be rearranged to give

$egin{displaymath} U= rac{1}{2}left[(k+k'),q_1^{,2} -2,k,q_1,q_2 + (k+k'),q_2^{,2} ight]. end{displaymath}$

Ofcourse it's not late to start aiming and preparing for IIT JEE kind of things at 11th standard. Just keep aiming 2wards the same and strive to excel in the same. take guidance from the experts and start to put ur effort in the same direction. Again i repeat the word hard work, dedication and strong will solve th purpose and shall help u in accomplishing ur desired goal.

All the very best for your future endeavours.

Just hard work, commitment and desire to succeed are the key ingredients to excel in any exam. I am sure you will come up with the flying colours.

Use the following identity to solve this integration

$\sin\left (x\right ) \sin\left (y\right ) = {\cos\left (x - y\right ) - \cos\left (x + y\right ) \over 2} \;$

As the water is leaking uniformly out of the bucket.

So the mass of the bucket is varying uniformly w.r.t height 'h'

If we plot the curve for force (along y-axis) versus height (or displacement along x-axis). Then area under the curve is the work done

Here area under the curve will be = (1/2) base*altitude

or W = (1/2)*(20m)*(20g-10g)N = 100g Joules = 1000 Joules

If b/a is the ratio of semi-major to semi-minor axis then we can calculate its eccentricity using the formula

$e = \sqrt{1 - \frac{b^2}{a^2}}$

The eccentricity is a positive number less than 1, or 0 in the case of a circle. The greater the eccentricity is, the larger the ratio of a to b, and therefore the more elongated is the ellipse. Moreover, the distance between the foci is 2ae.

Quantum Physics and classical physics

1) Quantum physics is applicable where classical mecanics/physics fails.

For example classical physics is applicable in classical or macro world, but it fails to explain the laws of physics in microworld like atomic structure, black body radiation etc.

2) Thus quantum physics a bridge to fill the gap of macro world and micro world where in the later case newtons laws are not sufficient to expalin the cause.

3) The basic difference b/w the two is that in quantum physics there is quantisation but in classical physics it is continuum or continuous spectrum.For example, in atomic physics derived from laws of quantum mechanics (SCHRODINGER EQUATION & DIRAC EQUATION) the energy levels of an electron that it can occupy in atom are discrete but not continuous and also the electron cannnot revolve in any arbitrary orbit. As only those orbits are allowed in which angular momemtum of an electron is integral multiple of h/2

(bhor's postulates). This is explained on the basis of quantum mechanics only whereas classical mechanics fails.

4) Phtoelectric effect is also explanable on the basis of quantum mecahnics.

5) Other examples of quantum physics are HEISENBERG'S UNCERTAINTY PRINCIPLE, PLANK'S LAW, DAVISON GERMER EXPERIMENT, COMPTON SCATTERING, FRANK HERTZ EXPERIMENT, STERN GERLACH EXPERIMENT etc.

6) Spin (one of the four quantum numbers) which is an intrinsic property of a particle can be expalined only on the basis of quantum physics using DIRAC'S theory whereas classical physics fails to explain the same.

7) But all the laws of quantum physics converges to classical mechanics for macro world.

If you mean Parallel and Perpendicular axes theorem for moments of Inertia, then answer is yes. These theorems are applicable to uniform as well as non uniform bodies.

Yes I agree upon uday_zingtudor's suggestion.

Remember it as rule that the for a given perimeter the area of Right angled triangle is maximum.

Since, eⁱ= cos

+ i sin

Therefore, e^i/2= cos

/2 + i sin

/2

or e^i/2= 0+ i*1 = i

or iⁱ = (e^i/2)ⁱ= (e^i*i/2) = e^{- /2}

or, iⁱ = e^{- /2}

When we are inside the room, or under the open sky, everytime we face atmospheric pressure and hence dont feel any difference inside or outside. It is only when we start moving about the surface of earth the altitutde increases and the column of atmosphere above us decreases, similar is the case when one goes down from the sea level.

Inside the room we feel atmospheric pressure becauseof pascal's law according to which when a pressure is applied on the surface of some fluid, the equal pressure is transferred through out the bulk or volume of fluid.The same concept is applicable in hydraulic lifts.

To obtain the circle, ellipse, parobola in polar equation form use the conversion of coordinates as follows.

Converting between polar and Cartesian coordinates

A diagram illustrating the conversion formulas.

The two polar coordinates r and ? can be converted to Cartesian coordinates by

$x = r cos heta ,$

$y = r sin heta. ,$

From the above two formulas, r and ? can be defined in terms of x and y:

$r = sqrt{x^2 + y^2} ,$

$heta = egin{cases} arctan( rac{y}{x}) & mbox{if }x>0 mbox{ and }yge0\ arctan( rac{y}{x}) + 2pi& mbox{if }x>0 mbox{ and }y<0\ arctan( rac{y}{x}) + pi& mbox{if }x<0\ rac{pi}{2} & mbox{if }x=0 mbox{ and }y>0\ rac{3pi}{2} & mbox{if }x=0 mbox{ and }y<0 end{cases}$

With this formula ? is obtained in [0, 2?), or [0°, 360°).

ONE IS ILLUSTRATED AS AN EXAMPLE FOR ELLIPSE

If an ellipse is not centered at the origin of an x-y coordinate system, but again has its major axis along the x-axis, it may be specified by the equation

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

where (h,k) is the center.

If an ellipse is not centered at the origin of an x-y coordinate system and has its major axis along the y-axis, it may be specified by the equation

$\frac{(y-k)^{2}}{a^{2}} + \frac{(x-h)^{2}}{b^{2}} = 1$

where (h,k) is the center.

In polar coordinates, where the origin is one focus of the ellipse:

$r = \frac{ a(1-e^{2})}{1 + e\cos(\theta)}$

A Gauss-mapped form:

$\left(\frac{a^2\cos\phi}{\sqrt{a^2\cos^2\phi+b^2\sin^2\phi}},\frac{b^2\sin\phi}{\sqrt{a^2\cos^2\phi+b^2\sin^2\phi}}\right)$

This is the question of limits when x tends to 1

So f{1}= lim_x0 {[log(1+x)-log2][3.4^x-1-3x]}/{(7+x)^1/3-(1+3x)^1/2}sinpix

To solve this apply L-Hospitals rule, according to which we differentiate the numerator and denominator and take the ratio then put the value of x as 1. keep on differentiating when until we don't get 0/0 or real/0 that is indefinite form.

That's the perfect solution provided by krish.

Take one salute for that

Uniform circular motion

The realm of physics consists of two types of circular motion: uniform circular motion and non-uniform circular motion.

Uniform circular motion describes motion in which an object moves with constant speed along a circular path.

Acceleration and velocity

Since the velocity is tangent to the circular path, no two velocities point in the same direction. Although the object has a constant speed, its direction is always changing. This change in velocity causes the object to accelerate. Similar to the velocity, the acceleration?s magnitude is held constant and the direction is forever changing. Such an object experiences a constantly changing acceleration pointing radially inwards (centripetally), which is perpendicular to its velocity. This acceleration is known as centripetal acceleration.

Image:Uniform circular motion diagram 0001.JPG

The magnitude of the acceleration is given by

a = v 2 / r

, where

v

is the speed of the object and

r

is the radius of its path, or

a = (4? 2 r) / t 2

, where

r

is the radius of the object and

t

is the time it takes the object to travel a distance.

How to derive these equations of acceleration

Image:Uniform circular motion diagram 2.JPG

Initial velocity = v₁, Final velocity= v₂,

a =

acceleration,

r =

radius of circular path,

s =

sector traveled

We follow the head-tail rule for adding vectors. Since the initial velocity points in the opposite direction as it is drawn on the circle diagram, we label it - v₁. Thus getting the equation: $\triangle v=$ v₂-v₁

a =

$(\triangle v)/t=$ (v₂-v₁)

/ t

a =

(v₂-v₁)

/ t

a t =

|v₂-v₁| We use the absolute value of v₂-v₁ because we only focus on the magnitude of the velocity.

At this point we can set up a proportion: |v₂-v₁|

/ v = s / r

We substitute

a t

for |v₂-v₁|, which gets us:

(a t) / v = s / r

(a r) / v = s / t

(a r) / v = v

,

s / t

becomes

v

because we have a distance over time which gives us velocity. However, this is only when we have

s

as a very small value. This is because as

s

gets smaller, the closer it is to the curvature of the circular path.

a = v 2 / r

, this is an equation of centripetal acceleration with respect to velocity because the radius remains constant.

We can derive the other equation by using the equation

v = d / t

. The distance of a circle is

2? r

. We can substitute

2? r

for

d

, which gets us:

v = (2? r) / t

. With this new information we can substitute

(2? r) / t

for

v

in the equation

a = v 2 / r

.

a = ((2? r) / t) 2 / r = (4? 2 r) / t 2

a = (4? 2 r) / t 2

, this is the equation of centripetal acceleration with respect to time because the radius remains constant.

Centripetal force

Image: Uniform_circular_motion_diagram_3.JPG

The acceleration is usually considered to be due to an inward acting force, which is known as the centripetal force. Centripetal force means ?center seeking? force. It is the force that keeps an object in its uniform circular motion. We determine this force by using Newton's second law of motion, F_net

= m a

, where F_net is the net force acting on the object (this is the centripetal force, F_c, of an object in uniform circular motion),

m

is the mass of the object, and

a

is the acceleration of the object. Since the acceleration of the object in uniform circular motion is the centripetal acceleration, we can substitute

v 2 / r

or

(4? 2 r) / t 2

for

a

. This gets us F_c

= (m v 2) / r

or F_c

= (4 m ? 2 r) / t 2

The centripetal force can be provided by many different things, such as tension (as in a string), and friction (as between a tire and the road).

An example of tension being the centripetal force is tying a mass onto a string and spinning it around in a horizontal circle above your head. The tension force is the centripetal force because it is the only force keeping the object in uniform circular motion.

Image: Uniform_circular_motion_tension.JPG

The

m

is the mass of the object, and the tension force is the centripetal force because it is keeping the object in uniform circular motion.

If a person were to cut the rope at one given point. The object would continue to move in the direction of the velocity.

Image: Uniform_circular_motion_tension_cut.JPG

As one can see, the string holding mass

m

is cut about ¾ of the way. After the string is cut, the tension force/centripetal force is no longer acting upon the object so there is no force holding the object in uniform circular motion. Therefore it continues going in the direction when it was last in contact with the force.

Similar to the tension force, the friction force between the tires of a car and the road is the centripetal force because it keeps the car moving in a circular path. If this were a frictionless plane, the car would not be able to move in uniform circular motion, and will instead travel in a straight line. Without the friction force acting upon it, no force is keeping the car in uniform circular, so it moves in a straight line.

Yes you are right in saying that there are 18 possible combinations.

Factors contributing towards rolling friction

Coupled oscillations

Spring-coupled masses

To obtain the circle, ellipse, parobola in polar equation form use the conversion of coordinates as follows.

Converting between polar and Cartesian coordinates

Uniform circular motion

Acceleration and velocity

How to derive these equations of acceleration

Centripetal force

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