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In this example. we are given that a body is released from rest from top of a fixed

spherical surface. such that it starts sliding down . we are required to find the acceleration

of body when it is at a position at an angle thirty degrees from the initial position.

let us first draw the, physical situation , we are given that, there is a spherical

surface . from the top of the spherical surface we are given, that body starts sliding down

it starts sliding down that means, if we consider its initial speed is zero , as its

come down , because the work done is by gravity so its speed will increase .

say the radius of this , surface is r , and when the body travels , up to a position,

where the radius makes an angle theta with the vertical or initial position , we can

say the body will have speed v , and in this situation, from initial to this position the

body fall by a distance h , and the value of h we can simply write as , r minus r coz

theta , so in this situation it will be r into one minus coz theta .

and by work energy equation we can say that whatever work is done by gravity, in moving,

the body from this point to this point that will be the gain in kinetic energy . so here

we can write, m g h , that is m g r into one minus coz theta equals to, half m v square

here m get cancelled out and the speed we will get as root of, two g r, one minus coz

theta . this we can get directly by writing v is equal

to root of two g h by energy conservation. we will get this equation.

now at this position, if we just see the normal reaction acting on it is n . it will experience

a, force , gravitational force in downward direction that is m g , this angle is theta

we are required to find the acceleration of body when it is at a position at an angle

thirty degrees from the initial position. so if theta is thirty degrees we can say , in

this situation, the normal reaction plus the centrifugal force on it is equal to m g coz

theta or m g coz thirty but we don't need to use those relations . if we talk about

acceleration at this point , we can simply say , normal acceleration. at point p. say

this point we denote by p . is. we know very well that normal acceleration

we write as v square by r , on substituting the values we get, two g r get cancelled out

it is one minus coz thirty degrees. so we will get the result as. coz thirty we

can write as root three by two, this can be written as two g one minus root three by two,

which can be written as g into two minus root three , this is the normal acceleration of

body at point p. and we can see that this weight is acting

on it in downward direction , and there will be a tangential component, i.e. m g sine theta

that will be experienced by the body along the tangent. so we can simply state that the

tangential acceleration, at point p is , if we talk about this body , this tangential

acceleration can be written as, m g sine theta up on m so this will be g sine theta .

now, if we put the value of theta that is thirty degrees so it comes out to be g by

two. and if we talk about the total acceleration we can state it as . total acceleration of

a body that is a-total can be written as root of a-tangential square plus a-normal square

when we substitute the values it will be root of

a tangential square is g square by four plus a normal square which is g square into two

minus root three whole square. on simplification, we get, this will be g, we can be simply common

out, and the remaining factor we will get is one by four plus two minus root three whole

square so this will be four plus three , minus two root three .

so this will give us the value, root of, one by four plus, four plus, three, minus four

root three. this will be the answer to this problem.

For more infomation >> 27. Physics | Circular Motion | Example-13 on Circular Motion | by Ashish Arora - Duration: 5:20.

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27. Physics | Circular Motion | Example-13 on Circular Motion | by Ashish Arora (GA) - Duration: 5:20.

In this example. we are given that a body is released from rest from top of a fixed

spherical surface. such that it starts sliding down. we are required to find the acceleration

of body when it is at a position at an angle thirty degrees from the initial position.

let us first draw the, physical situation,

we are given that, there is a spherical

surface. from the top of the spherical surface we are given, that body starts sliding down

it starts sliding down that means, if we consider its initial speed is zero, as its

come down, because the work done is by gravity so its speed will increase.

say the radius of this, surface is r, and when the body travels, up to a position,

where the radius makes an angle theta with the vertical or initial position, we can

say the body will have speed v, and in this situation, from initial to this position the

body fall by a distance h, and the value of h we can simply write as, r minus r coz

theta, so in this situation it will be r into one minus coz theta.

and by work energy equation we can say that whatever work is done by gravity, in moving,

the body from this point to this point that will be the gain in kinetic energy. so here

we can write, m g h, that is m g r into one minus coz theta equals to, half m v square

here m get cancelled out and the speed we will get as root of, two g r, one minus coz

theta this we can get directly by writing v is equal

to root of two g h by energy conservation. we will get this equation.

now at this position, if we just see the normal reaction acting on it is n. it will experience

a, force, gravitational force in downward direction that is m g, this angle is theta

we are required to find the acceleration of body when it is at a position at an angle

thirty degrees from the initial position. so if theta is thirty degrees we can say, in

this situation, the normal reaction plus the centrifugal force on it is equal to m g coz

theta or m g coz thirty but we don't need to use those relations. if we talk about

acceleration at this point, we can simply say, normal acceleration. at point p. say

this point we denote by p. is we know very well that normal acceleration

we write as v square by r, on substituting the values we get, two g r get cancelled out

or it is one minus coz thirty degrees. so we will get the result as. coz thirty we

can write as root three by two, this can be written as two g one minus root three by two,

which can be written as g into two minus root three, this is the normal acceleration of

body at point p. and we can see that this weight is acting

on it in downward direction, and there will be a tangential component, i.e. m g sine theta

that will be experienced by the body along the tangent. so we can simply state that the

tangential acceleration, at point p is, if we talk about this body, this tangential

acceleration can be written as, m g sine theta up on m so this will be g sine theta.

now, if we put the value of theta that is thirty degrees so it comes out to be g by

two. and if we talk about the total acceleration we can state it as. total acceleration of

a body that is a-total can be written as root of a-tangential square plus a-normal square

when we substitute the values it will be root

a tangential square is g square by four plus a normal square which is g square into two

minus root three whole square. on simplification we get, this will be g, it can be taken

out, and the remaining factor we will get is one by four plus two minus root three whole

square so this will be four plus three minus two root three.

so this will give us the value, root of, one by four plus, four plus, three, minus four

root three. this will be the answer to this problem.

For more infomation >> 27. Physics | Circular Motion | Example-13 on Circular Motion | by Ashish Arora (GA) - Duration: 5:20.

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4. Physics | Rigid Body Dynamics | A Sphere rolling on a Wedge | by Ashish Arora (GA) - Duration: 5:31.

In this illustration we'll analyze sphere rolling on a wedge. we are given that a uniform

solid sphere of mass m and radius r. rolling without slipping down on an inclined wedge

of same mass as shown. here we are required to find the velocity of wedge, when sphere

reaches the bottom of this wedge. now, in this situation we can analyze, the case when

the sphere reaches the bottom of wedge. then wedge will attain a leftward velocity because

in this situation no external force is acting on the system in horizontal direction so its

system center of mass must be at rest here the wedge is of mass m and the sphere is also

of mass m here we can consider if this sphere attains a velocity v 1 with respect to be

wedge and wedge is moving toward left with a velocity v 2 with respect to ground then

here we can write as no external force in horizontal direction is acting on system here,

we use the conservation of momentum in horizontal direction we use by conservation of momentum

as here we can right toward left the wedge is attained a momentum m v 2 and toward right

the ball is attain a momentum m this v 1 coz theta minus v 2 because here we have taken

the v 1 as velocity of ball with respect to wedge so in this situation with respect to

horizontal v 1 is moving at an angle theta so ball is moving toward right at v 1 coz

theta as well as it is moving with the wedge toward left with v 2 so simplifying this relation

here m gets cancelled out and this will give us v 1 is equal to 2 v 2 by coz theta and

we can write by, work energy theorem the kinetic energy gain in wedge as well as in that of

ball is due to the work done by gravity so here work done by gravity is m g h and that

should be the gain in kinetic energy of wedge plus ball so for wedge it is half m v 2 square

plus for ball we can write it is half m it has 2 velocity components so we can write

1 as the horizontal velocity component which is v 1 coz theta minus v 2 whole square plus

v 1 sine theta whole square plus the rotational energy attained by the ball as it is rotating

at angular speed omega which is with respect to wedge its velocity is v 1 divided by its

radius. so we can write its rotational energy as half i omega square which is 2 by 5 m r

square multiplied by v 1 by r whole square now if we simplify this relation here this

gives us m gets cancelled out in the whole equation this gives us 2 gee h is equal to

v 2 square plus v 1 square plus v 2 square minus 2 v 1 v 2 coz theta plus 2 by 5 v 1

square and now for v 1 we can substitute the value in terms of v 2 from this equation this

gives us 2 gee h is equal to here this will be 2 v 2 square plus v 1 square plus 2 by

5 v 1 square we can write as 7 by 5 v 1 square so this is 7 by 5 and v 1 square we can write

as 4 v 2 square divided by coz square theta and minus here we can write v 1 coz theta

is equal to 2 v 2 so this becomes 4 v 2 square further on simplifying this relation we get

the value of v 2 is equal to under the root of 5 gee h divided by 14 sek square theta

minus 5 that is a result of this problem.

For more infomation >> 4. Physics | Rigid Body Dynamics | A Sphere rolling on a Wedge | by Ashish Arora (GA) - Duration: 5:31.

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23. Physics | Kinematics | Maximum Range of Projectile From a Tower | by Ashish Arora - Duration: 4:59.

in this illustration, we'll study about the maximum range of a projectile which is

thrown from a tower. we are given that from top of a tower of height h, a ball is thrown

at a speed u. we are required to find the angle of projection at which the range of

ball in ground is maximum. and if we draw the situation it is like this, on ground from

a tower of height h. a ball is thrown with the speed, u at an angle theta. follows the

trajectory and finally hit the ground. say this range is r, so we are required to find

this value of theta for which the value of r is maximum. so here, we can use equation

of trajectory. where we use this direction as x axis, and vertically upward as y axis

with the initial point as origin. so trajectory equation is, y is equal to x tan theta minus.

g x square by 2 u square coz square theta. so here, we can see. if we substitute the

y is equal to minus, h we get the range as r so this point will be having coordinates

r comma, minus h, so here we can write minus h is equal to, r tan theta minus. g r square

by 2 u square, coz square theta. in this expression. we can write for maximum, arc. the value of

d r by d theta should be equal to zero. so if we differentiate it, this lefthand side

becomes zero and right hand side here, if we differentiate it. this will become tan

theta multiplied by d r by d theta, plus, r, seck square theta. here i have differentiate

it this term by using product rule, minus here also i can use, g r square by 2 u square

seck square theta. so i can take g r square by 2 u square. and seck square theta when

we differentiate this will give us 2 seck theta multiplied by seck theta tan theta.

and further minus, we can differentiate this r square. so this will be, g by 2 u square.

seck square theta multiplied by we can write it 2 r d r by. d theta. so, this the expression

we are getting after differentiating. and as, if we put d r by d theta equal to zero.

this term, and this term will vanish. and here we are getting. r seck square theta is

equal to. g r square by 2 u square multiplied by 2 seck square theta. tan theta. here, 1

r again gets cancelled out 2 also gets cancelled out, and the seck square theta will also gets

cancelled out and we keep in mind that r can not be zero seck square theta can not be zero.

so, from here we are getting the value of r is equal to u square by, g tan theta. so

in this situation, if we substitute the value in equation 1. we can see, from, equation

1 we are getting. this as minus h is equal to if we substitute the value of r is u square

by g tan theta here tan theta gets cancelled out this u square by g, minus, this. g by,

2 u square coz square theta multiplied by r square. so if we square it this will be

u to power 4 by, g square, tan square theta can be written as sine square theta by coz

square theta. so here when g gets cancelled out, coz square theta gets cancelled out and

1 u square also gets cancelled out. so the result we are getting after simplification

as u square by g minus, h. is equal to u square by 2 g sine square theta. so on simplifying

this we are getting the value of sine theta is equal to root of, u square by, twice of

u square minus, g h. or the value of theta we are getting here is sine inverse of, root

of u square by twice of u square minus, g h. this is the final result of our problem.

the angle at which when, the ball is to projected from this point so that the maximum range

on ground is obtained.

For more infomation >> 23. Physics | Kinematics | Maximum Range of Projectile From a Tower | by Ashish Arora - Duration: 4:59.

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23. Physics | Kinematics | Maximum Range of Projectile From a Tower | by Ashish Arora (GA) - Duration: 4:59.