@AbhasKumarSinha nope. But it's somehow in between bs and 12th. Spend your time on them. They'll look easy after ample practice. You can learn some extra methods must be very helpful. Like langrangian mechanics or so . These will help. Although they are optional.
I'm not sure how to make the virtual work idea intuitive. I guess I learned it so long ago that it feels obvious. The best course is probably to just accept that it works.
Essentially it's a force balance. Suppose you have some system where there are two forces acting. In the case of the thread one force is gravity and the other is the tension in the thread.
Start with the simple case where the two forces point in opposite directions i.e. along the same line.
If we move the contact point of the forces by a small distance $dx$ then one force does work $F_1dx$ and the other force does work $-F_2dx$. Yes?
We have two forces, $F_1$ and $F_2$, shown by the red and blue arrows, acting in opposite directions at the point shown by the left dashed line. These could be any two forces e.g. two springs, a string lifting a weight, whatever. I'm making a general argument that doesn't depend on exactly what the two forces are caused by.
Now suppose the point where the two forces act moves by a distance $dx$, as shown in the lower diagram. The forces do work $F_1dx$ and $F_2dx$ (they will have opposite signs because the forces are in opposite directions)
@Abcd I'm not sure I can do that. You'll just have to accept that it applies to all systems. In the case of the thread I displaced the thread down the hemisphere by a distance $dx$, calcuated the work done by (a) gravity and (b) the tension in the thread, added them together and set the total to zero. That then gives the tension in the thread.
So it's exactly the same principle as my simple example above.
But I no longer remember how to prove that the principle applies to all cases.
Because the thread is on a sphere, when you displace the thread down by $dx$ the radius of the circle it forms increases. Since the radius increases the circumference increases, and the circumference is just the length of the thread.
With some simple geometry you can work out how much the circumference increases - call this $dC$.
Then if the tension in the thread is $T$ the work done by stretching the thread by a distance $dC$ is just $TdC$.
And the virtual work condition says that if the two forces, gravity and thread tension, are in balance then:
$$ mg dx = T dC $$
Or rearranging this:
$$ T = mg \frac{dx}{dC} $$
And $C = 2\pi R$ so $dx/dC = dx/(2\pi dR)$ giving us:
You can either use some elementary geometry to work out $dx/dr$ or as I did just spot that it is the gradient of the tangent to the hemisphere at the thread.
@Abcd Cool. I need to get back to work now for about an hour.
@Hema the surface energy is just proportional to the surface area, is the surface energy constant when you split a big bubble into smaller bubbles? I would have guessed not.
Maybe it's because the pressure inside a bubble is inversely proportional to thr bubble radius, so when you split a bubble in two the two smaller bubbles have a smaller total volume.
@Nobodyrecognizeable Force applied to hollow sphere : Equations (2) and (3) are inconsistent. (2) assumes friction is backward, (3) assumes it is forward, giving opposite torque to F. Abhas is correct : for h<2R/3 friction is backward but for h>2R/3 it is forward.
@Jasmine Rutherford Scattering Experiment : Alpha particle and nucleus are both +ve so PE is always +ve. And of course KE is always +ve. Think of alpha particle colliding head-on and bouncing backwards : KE is high and PE low when alpha particle is far from nucleus, KE is low (zero at closest point) and PE high when alpha particle is near nucleus. It is the same if alpha particle passes the nucelus : KE is highest far from the nucleus.
If incident particle and nucleus had opposite signs PE would always be -ve. Then KE increases closer to nucleus (more +ve) and PE would decrease (more -ve).
@Abcd in pure rolling there is never any relative motion between the rolling object and the surface it's rolling on i.e. if you take any moment in time the bit of the rolling object in contact with the surface is stationary relative to the surface.
@Abcd Suppose the surface is completely flat, and we'll ignore air resistance. Then there are no forces acting on the rolling object. It will roll in a straight line at constant speed forever. Since the torque on the rolling object is zero the force between the rolling object and the surface is zero. Even with a perfectly frictionless surface the object just keeps (pure) rolling.
Now suppose the surface is not horizontal but has a slope. We'll assume the surface is still frictionless. Then the linear velocity of the object reduces as it moves up the surface. The deceleration is just $g\sin\theta$, where $\theta$ is the angle of the surface to the horizontal.
That means the linear velocity $v$ is reducing, but because the surface is frictionless the angular velocity remains unchanged so $r\omega > v$. The object must now start slipping, and the slipping velocity increases as the object moves up the slope.
The only way the object can avoid slipping is if the slope can exert a force on the object to change its angular velocity. For no-slipping that force has to be large enough to make $d(r\omega)/dt = dv/dt$.
@JohnRennie if a rolling object (not necessarily pure rolling) goes from bottom to the top of a curved surface like a hill/parabola will it be at rest on its maximum height?
@Abcd Perhaps resolve the (horizontal) velocity of the balls along the string, then resolve the velocity of the string along the vertical. So if $\theta$ is the angle between the string and the horizontal and $v$ the velocity of either ball then $v\cos\theta\sin\theta=v_0$.
@Abcd Ah yes. I misread the question. The solution linked by John Rennie makes sense. Use a frame in which the midpoint is stationary and the balls each have velocity $v_0$ downwards. Then by conservation of angular momentum they still have velocity $v_0$ at all other positions, although no longer directed downwards.
I thought it was asking for the speed with which the balls move towards each other.
@Abcd In the ground frame the midpoint of the string moves upward with a constant velocity $v_0$. In the inertial frame of the midpoint the midpoint is stationary while the balls initially move downward with velocity $v_0$.
Angular momentum is conserved, and the length of the string each side of the midpoint remains the same, so the speed of the balls remains $v_0$ as they rotate about the midpoint in opposite directions.
When the balls are separated by $\ell/2$ they still have velocities $v_0$ but not directed towards each other. Instead directed $30^{\circ}$ below the horizontal.
@Abcd I've forgotten we need to transform back to the ground frame of reference by adding $v_0$ upwards. But this gives an equilateral triangle, so in ground frame the velocity of each ball is still $v_0$ directed at $30^{\circ}$ above the horizontal.
@Abcd What is your reasoning? That both cases are equivalent?
When you step off you continue moving tangentially. You continue to have AM about the centre, the AM of the disk itself does not change, so its rotational speed does not change.
Whereas when you step onto the disk you start with no AM about the centre.
A similar question is sand dropping onto or off a moving truck. When it drops off it continues moving forward with the truck, so the speed of the truck does not change. Whereas when it drops onto the truck it starts at rest but the truck is moving, it increases the mass of the truck but adds no linear momentum. The total momentum of the truck and contents does not change, but the mass increases so the velocity is reduced.
@Abcd Angular momentum about O is constant in both magnitude and direction (vertical). But angular momentum about C is directed always perpendicular to the string, and changes direction.
@Abcd Friction during rolling : depending on the situation it could be kinetic, static, or neither! If there is pure rolling (no sliding at the contact point) it is static if the object is accelerating and no friction if acceleration is zero.
xNow my book says that in case there is friction, whether or not F is greater than limiting value the elongation will remain same. Why is this so? How is this possible when the tension on each element dx will change in either case?
Now my book says that in case there is friction, whether or not F is greater than limiting value the elongation will remain same. Why is this so? How is this possible when the tension on each element dx will change in either case?
@LoopBack This is getting to sound like my explanation of using a frame in which the point at which the force is applied is at rest and the balls start with velocity $v_0$ downwards.
@Abcd Ah yes. I misread the question. The solution linked by John Rennie makes sense. Use a frame in which the midpoint is stationary and the balls each have velocity $v_0$ downwards. Then by conservation of angular momentum they still have velocity $v_0$ at all other positions, although no longer directed downwards.
So do you understand that the ball will move with a constant velocity as no force is acting after the impulse is given to it
Now look at this figure
Both the mid point of the string as well as the ball are moving with constant velocity. In the final position both have travelled equal distance =L/2 in equal time. Which means both have same velocity
That is $v_0$
@sammygerbil do you understand what I m trying to say.
@LoopBack Sorry I am still not getting it. Also, your diagram seems to show the balls touching in the final position, whereas the question says their separation is $\ell/2$.
If the balls were given an impulse vertically and the strings are not there then the balls will continue moving in parallel directions vertically and will never meet. The tension in the string is needed to change their direction and bring them together.
@LoopBack The balls are initially rotating about the midpoint of the string.
Balls are not given impulse vertical rather diagonally, see this
V
The impulse in the vertical direction will be counteractes by the gravitational force and the balls will remain on the ground
Now the only impulse left is the horizontal one which is used to initiate the motion.
@sammygerbil let me start from the basics, do see any horizontal external force on the balls. The answer is NO. So there won't be any tension to counter act it
@sammygerbil try to forget how you have approached the solution to this question. Because it isn't letting you to think differently. Think about it deeply
@sammygerbil Consider this situation. Two men 10m apart are holding a rope and starts running simultaneously with same velocity in the line join them. Will the be tension in the rope. The answer is NO
@LoopBack I think I understand what you mean but I am still confused. (My solution does not apply - I think - because I assumed there was no floor and no gravity.)
@Hema Probably the easiest explanation is to use a non-inertial frame of reference in which the block is stationary. If it is accelerating then there is an inertial force acting to the left on each segment. If there is friction there is a friction force acting on the left on each segment. Whatever the value of acceleration, the total forces to the left always balance the forces to the right.
The tension in each segment is the same as in the calculation without friction, so the elongation is the same.
@Hema Even when $F$ is less than limiting friction and the block does not move, the forces on each segment of the block must be balanced. The difference between tensions on left and right of each segment equals the friction force to the left on that segment.
When the block accelerates (to the right) the difference in these tensions equals the mass of the segment times its acceleration, which is equivalent to an inertial force to the left.
@Hema Another way to look at it is to hang the block vertically in a gravitational field. The force $F$ at the support is the total weight of the block. There is no acceleration. The difference in tensions pulling up and down on each segment equals the weight of the segment.
This is equivalent to the no-friction case. There is acceleration $a=F/m=g$. If there is friction then like gravity it applies the same downward force on each segment. If we keep the support force $F$ constant then to apply an extra friction force on each segment we must reduce the strength of gravity $g$.
The extreme case is that we reduce gravity to zero so that the only downward force is friction. There is no "acceleration" so the friction is static friction. ... In the horizontal case the "strength of gravity" is the acceleration $a$ which automatically reduces if we increase the friction force $f$ : $F-f=ma$.
@Hema Friction is downwards because the support force is pulling upwards, and friction opposes it, just as in the horizontal case the applied force $F$ acts to the right and friction $f$ acts to the left. Whether the block moves or not, $F$ attempts to move it to the right, so friction opposes this and acts to the left.
@Hema A downward friction force increases the support force $F$ required to keep the block balanced. If $F$ is fixed then we have to reduce the strength of gravity $g$ at the same time as we increase the friction force, in order to avoid changing $F$.
@sammygerbil But for each segment we have a downward weight of segment as well as a net upward support force F right? Why does friction oppose F and not weight?
@sammygerbil never mind I had got confused
@sammygerbil okI actually still am confused on that point
By F are we referring to the weight or the net upward tension in resonse to the weight
@Hema $F$ is the upwards external force on the block applied by the support to which it is attached. The block is in equilibrium so when there is no friction $F=mg$ the weight of the block (which is downwards).
When there is friction (which is applied downwards opposite to the direction of $F$) then $F=mg+f$.
@Hema Friction is opposite to $F$ in your problem when the block is horizontal. That is why we apply friction downwards in the vertical case, so that it is again opposite to the direction of $F$.
@LoopBack If you think it would be useful to others you could post a question and answer in Stack Exchange Meta, after checking there is not a solution there already. Or you could answer one of the open questions. See meta.stackexchange.com/search?q=mathjax+android
@Hema In the horizontal case there are two forces opposing $F$ : friction force $f$ and the inertial force $ma$. In the non-inertial frame of reference in which the block is stationary the inertial force $ma$ acting to the left replaces the mass x acceleration $a$ to the right.
A block with a pendulum slides down an incline. Regardless of the angle of the incline (and the acceleration down it) the angle of the pendulum is determined by the coefficient of friction rather than the angle of the incline.
