In the given figure, all surfaces have equal coeff. of restitution. The ball collides with the vertical wall at the highest point in its path. The path of the ball is shown.
sir my class sir posted me following solution of insect problem my mail.it is as follow....... Consider a right handed rectangular coordinate system and make the following assumptions: (i) The triangle is an equilateral triangle. (ii) The triangle lies on the yz-plane. (iii) The triangle hangs on the x-axis with the small ring at the origin O of the coordinate system. Then, each interior angle of the triangle is a 60° angle, so that the length of every side is (1) ..... Lr = L /sin 60° = 1.155 * L Assume for the moment that the insect remains stationary at the left vertex A of the triangle (the right vertex being at B). Then, the weight of the insect exerts a torque about O, which will cause the triangle to rotate counter clockwise. As the triangle turns about O, the insect falls down reducing its height relatve to some point and thereby reducing its gravitational potemtial energy (GPE) relative to that point. If we let GPE = 0 at the point where the left side of the triangle lies along the vertical (OA is along the vertical while AB, the lowest side of the triangle, makes an angle of 30° with the horizontal), then the total energy of the insect at that point is purely kinetic energy. On the other hand, at the start when AB is along the horizontal (and OA makes an angle of 30° to the left of the verticall), the total energy is purely gravitational potential. Then, after A reaches its lowest point, A will continue to move up again to the right until OA makes another 30° angle to the right of the vertical, conserving the total energy in the process. After that, A will fall back down again to the other side, transforming the gravitational potential energy to kinetic energy once more. In this way, a simple harmonic motion is produced by the torque about O due due to the weight of the insect at A at the start. The period of oscillation produced is, of course (2) ..... Pe = 2*π*sqrt ( Lr / g ) Now, let us suppose that the insect remains stationary at A long enough until OA makes an angular displacement of 60°. Then, when OA is about to make its first swing to the left, the insect must crawl to the right fast enough with speed U until it reaches B at the exact time (half the period PE) when the side AB is horizontal again (and the insect at the right vertex B for the first time). That means Lr = U*(0.5)*[ 2*π*sqrt ( Lr / g ) ] which gives (3) ..... U = Lr / [ π*sqrt ( Lr / g ) ] = (1 / π)*sqrt (Lr*g) When OA is about to swing to the right for the second time, the insect must crawl to the left with speed U given in (3). By the time AB makes a 30° angle to the right of the vertical again, the insect will be at the left vertex of the triangle once more. In moving this way, with speed U in (3) in the direction opposite to the direction of swing of the triangle, the insect can remain at the same point in space to the right of the vertical with always the same distance (4) ..... H = Lr – L = L* [ 1.155 - 1 ] = (0.155) * L according to (1), above the lowest position of either A or B.
A ladder of length L and uniform mass density stands on a frictionless floor and leans against a frictionless wall. It is initially held motionless, with its bottom end an infinitesimal distance from the wall. It is then released, whereupon the bottom end slides away from the wall, and the top end slides down the wall . When it loses contact with the wall, what is the horizontal component of the velocity of the center of mass?
i have a doubt in the above question is that what is the kinetic energy of center of mass when the end of the ladder which slides initially with the wall makes an angle theta with the wall...... please give detail of the kinetic energy .please bhaiya
the answer given in the solution is mr^2(dTheta/dt)^2+1/2I(dTheta/dt)^2 here m is the mass of the ladder and r is the distance of the center of mass from the corner formed by meeting floor and wall.Here theta is the angle between the wall and the ladder.It is also the angle between the position vector of the COM from the corner(which is mentioned above)and the wall.Her r=L/2, where L is the length of ladder.
sir my class sir posted me following solution of insect problem my mail.it is as follow....... Consider a right handed rectangular coordinate system and make the
ReplyDeletefollowing assumptions:
(i) The triangle is an equilateral triangle.
(ii) The triangle lies on the yz-plane.
(iii) The triangle hangs on the x-axis with the small ring at the origin O
of the coordinate system.
Then, each interior angle of the triangle is a 60° angle, so that the length
of every side is
(1) ..... Lr = L /sin 60° = 1.155 * L
Assume for the moment that the insect remains stationary at the left
vertex A of the triangle (the right vertex being at B). Then, the weight of
the insect exerts a torque about O, which will cause the triangle to rotate
counter clockwise. As the triangle turns about O, the insect falls down
reducing its height relatve to some point and thereby reducing its
gravitational potemtial energy (GPE) relative to that point. If we let
GPE = 0 at the point where the left side of the triangle lies along the
vertical (OA is along the vertical while AB, the lowest side of the triangle,
makes an angle of 30° with the horizontal), then the total energy of the
insect at that point is purely kinetic energy. On the other hand, at the start
when AB is along the horizontal (and OA makes an angle of 30° to the left of
the verticall), the total energy is purely gravitational potential. Then, after A
reaches its lowest point, A will continue to move up again to the right until
OA makes another 30° angle to the right of the vertical, conserving the total
energy in the process. After that, A will fall back down again to the other side,
transforming the gravitational potential energy to kinetic energy once more.
In this way, a simple harmonic motion is produced by the torque about O due
due to the weight of the insect at A at the start. The period of oscillation
produced is, of course
(2) ..... Pe = 2*π*sqrt ( Lr / g )
Now, let us suppose that the insect remains stationary at A long enough
until OA makes an angular displacement of 60°. Then, when OA is about to
make its first swing to the left, the insect must crawl to the right fast enough
with speed U until it reaches B at the exact time (half the period PE) when
the side AB is horizontal again (and the insect at the right vertex B for the
first time). That means Lr = U*(0.5)*[ 2*π*sqrt ( Lr / g ) ]
which gives
(3) ..... U = Lr / [ π*sqrt ( Lr / g ) ] = (1 / π)*sqrt (Lr*g)
When OA is about to swing to the right for the second time, the insect must
crawl to the left with speed U given in (3). By the time AB makes a 30° angle
to the right of the vertical again, the insect will be at the left vertex of the
triangle once more.
In moving this way, with speed U in (3) in the direction opposite to the
direction of swing of the triangle, the insect can remain at the same point
in space to the right of the vertical with always the same distance
(4) ..... H = Lr – L = L* [ 1.155 - 1 ] = (0.155) * L
according to (1), above the lowest position of either A or B.
where is the mistake in my sir solution???? please tell bhaiya
ReplyDeletein the problem i posted, the frame does not rotate at all...
ReplyDeleteIn your solution, you have assumed the insect to be 'falling' and applied some cons. of eng...
but if the insect moves as described in my solution, the frame is always at equilibrium..
is the frame is not at equilibrium then is the SIR solution is correct????
ReplyDeleteif the frame is not at equilibrium then is the SIR solution is correct???? please tell bhaiya
ReplyDeleteis the answer of this problem e=1/2
ReplyDeleteyup...coefficient of restitution is half
ReplyDeleteif the frame is not at equilibrium then is the SIR solution is correct???? please tell bhaiya
ReplyDeletee=1/2 bhaiya plz see the last ques also i have i doubt in that
ReplyDeletethis was an easy one
ReplyDelete1/2
or -1(which is not possible)
A ladder of length L and uniform mass density stands on a frictionless floor and leans against a frictionless wall. It is initially held motionless, with its bottom end an infinitesimal distance from the wall. It is then released, whereupon the bottom end slides away from the wall, and the top end slides down the wall . When it loses contact with the wall, what is the horizontal component of the velocity of the center of mass?
ReplyDeletei have a doubt in the above question is that what is the kinetic energy of center of mass when the end of the ladder which slides initially with the wall makes an angle theta with the wall...... please give detail of the kinetic energy .please bhaiya
ReplyDeletethe answer given in the solution is mr^2(dTheta/dt)^2+1/2I(dTheta/dt)^2 here m is the mass of the ladder and r is the distance of the center of mass from the corner formed by meeting floor and wall.Here theta is the angle between the wall and the ladder.It is also the angle between the position vector of the COM from the corner(which is mentioned above)and the wall.Her r=L/2, where L is the length of ladder.
ReplyDeletehere I=(1/3)mL^2
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteAmritpal, can u kindly stop running for too many extra questions, you should do only things which give u confidence
ReplyDeleteThis comment has been removed by the author.
ReplyDeletehi piyush!
ReplyDeletei'll suggest that if you haven't used instantaneous axis concepts till now (which i call COR), then you better continue with the same..
and Amritpal yaar, i did'nt get what you sir concluded..i had asked for the nature of the motion..but otherwise, he was correct..
the ans. to this one was 1/2..amritpal..i'll think abt. you problem..
ReplyDeletehow can we solve the question given in physics workshop(mechanics-2) qus -61,62,63 with help of COR theorem
ReplyDeletePlease give the solution to this problem,.
ReplyDelete