The ball on a sphere problem

Conversation started Feb 7, 2019 at 18:51.

tatan

Feb 7, 2019 18:51

if only our country provided opportunity in all fields. You want to study engineering physics but theres no surety of what will happen after 4 years ;-(

7 hours later…

Zerix

Feb 8, 2019 02:04

Zerix

Feb 8, 2019 02:38

@Pranav I guess because there is no torque at the point of contact so angular momentum is conserved. Since Moment of inertia decreases so onegs increases

Which further implies linear velocity increases

@Don'tbeax-tripledot can you post the solution in your free time pls

2 hours later…

Zerix

Feb 8, 2019 05:05

@JohnRennie good morning

John Rennie

@Zerix morning :-)

Zerix

@JohnRennie are you free for giving hint

John Rennie

Yes, I'm around for several hours now

Zerix

Okay

@JohnRennie A small ball of mass $m$ lies on the superior pole of a frictionless semi-sphere of mass $M$ which lies on a horizontal frictionless surface. The ball is smoothly released from its unstable equilibrium position and starts moving on the surface of the semi-sphere (with no initial velocity). Find the angle $\theta$ for which the ball detaches from the semi-sphere. Solve the problem in the particular case $m=M$.

How to proceed... COM frame?

John Rennie

That doesn't sound like a JEE problem ...

Zerix

Feb 8, 2019 05:09

It's not. But still this could be solved by jee techniques?

Do we need to make diffrential equation here

John Rennie

I would work in the COM frame and use conservation of energy.

The momenta of the ball and hemisphere have to be equal and opposite and their total KE must add up to $mgh$, where $h$ is the vertical distance the ball has fallen.

Zerix

@JohnRennie okay

John Rennie

Assume the ball stay in contact, so you know the direction of motion of the ball because it's the tangent to the hemisphere.

Zerix

Yes

John Rennie

Then you can calculate the normal force, and the point at which it falls below zero.

Zerix

Feb 8, 2019 05:16

@JohnRennie N=mv^2/r+ mgcos(thetha)

I feel this is incorrect

Wait at the point of leaving contact

mgcos(thetha) = mv^2/r

Is this what you were talking about?

John Rennie

The falling sphere doesn't trace out an arc of a circle because the hemisphere is accelerating in the opposite direction, so the acceleration won't simply be v^2/r. I'd have to sit down and start working in the problem to figure out what the normal force is.

Zerix

@JohnRennie better to leave then

@JohnRennie physics.stackexchange.com/questions/262282/…

Does anyone answer given there matches the step to follow

Nobody recognizeable

Feb 8, 2019 05:31

@JohnRennie are you free now?

John Rennie

@Zerix that's a stationary hemisphere i.e. the particle is sliding on a fixed surface. Your problem is harder because the hemisphere is allowed to slide as well.

@Nobodyrecognizeable hi, yes, I'm around for a while.

Nobody recognizeable

@JohnRennie as the hemisphere is a frictionless surface so it doesn't pose any problem. Do it the normal way.

@JohnRennie anyway wanted to show you some programmes of c.

John Rennie

OK ...

Nobody recognizeable

@JohnRennie you know what the programme wants to do. I just want to confirm if it gives the output or not.

John Rennie

It asks the user for a number, call this n, then it reads n-1 more numbers from the user. That's probably a bug - I'd guess it's supposed to read n more numbers.

Then it prints out the numbers the user entered, but for some reason it keeps printing the first number.

Nobody recognizeable

Feb 8, 2019 05:38

@JohnRennie it wants you to put a number then print the first and last integer of the number.

The number should be entered via enters.

John Rennie

You means that was the problem set? If so that code doesn't work.

Nobody recognizeable

@JohnRennie should the return 0 be the problem?

John Rennie

"return 0" ?

Nobody recognizeable

@JohnRennie yep that's the question.

@JohnRennie you are doing it in c right?

John Rennie

The first task is to prompt the user for a number. The line scanf("%d", &n) does this. It accepts a number entered by the user and stores it in the variable n. So far so good.

Now you have to print the first and last digit of the number. I can think of two ways to do this. On way would be to take the modulus 10 of the number i.e. n % 10. This will give you the last digit. OK so far?

Nobody recognizeable

Feb 8, 2019 05:45

@JohnRennie sorry one phone call came . As far i know n%10 means the leftout after dividing n by 10.

John Rennie

Correct, and that is going to be the last digit of the number. For example if the number is 1234 then the remainder after dividing by 10 is 4 i.e. the last digit.

Nobody recognizeable

@JohnRennie ok . Thats brilliant. But what to do with the first number.

John Rennie

Keep dividing the number by 10 until it is less than 10. Something like:

while (n > 9)
  n = n/10;

Take the example of 1234 again. Dividing by 10 gives 123, then 12, then 1 and now we stop because 1 < 10. What we have left is the first digit i.e. "1".

Nobody recognizeable

@JohnRennie wouldn't it give 1.234?

John Rennie

Integer division always returns an integer, so the fraction part is dropped.

e.g. with integer division 3/2 = 1 because integers have to be whole numbers so the .5 is dropped.

Nobody recognizeable

Feb 8, 2019 05:53

I see . So it would need three variables ie int i,j,k. Ok i am trying to type it in mobile c. If i run into problems I'll be back again. Bye for now.

John Rennie

You don't need to include math.h and conio.h. Just include stdio.h.

Don't be a x-triple dot

@Zerix For the semi-sphere problem? Sure, will do.

Nobody recognizeable

@JohnRennie shouldn't this do the job?

John Rennie

while (0<j<10) isn't valid C.

Nobody recognizeable

@JohnRennie ah ok.

John Rennie

Feb 8, 2019 06:06

#include <stdio.h>

int main(int argc, char* argv[])
{
  int input_number, first_digit, last_digit;

  printf("Enter a number: ");
  scanf("%d", &input_number);

  last_digit = abs(input_number) % 10;

  first_digit = abs(input_number);
  while (first_digit > 9)
    first_digit /= 10;

  printf("First_digit = %d, last digit = %d\n", first_digit, last_digit);

  return 0;
}

That's what I did ...

Nobody recognizeable

@JohnRennie hell that even runs in bad compiler like mobile c. Do you find any inaccuracies in my version of the code?

John Rennie

The line n%10==i doesn't do what you want. == is the equality test not the assignment operator. That line compares n%10 to i and returns true if they are equal or false if they are different.

Nobody recognizeable

@JohnRennie what if i did n%10=i?

John Rennie

@Nobodyrecognizeable = sets the left side equal to the right side. So that line is trying to set n%10 equal to i.

You want i = n % 10

Nobody recognizeable

@JohnRennie but there's still problem with the first integer.

It shows up to be 0. Upon typing 678.

Abcd

Feb 8, 2019 06:16

@JohnRennie do you know file handling in java?

John Rennie

@Abcd I've never done it, but Java is similar enough to C++ that I expect I could help ...

@Nobodyrecognizeable you need to first set j equal to n, then start dividing j by 10. At the moment you aren't setting j to anything, so it is defaulting to zero.

Also you need to loop while j>9, not while j<10.

Nobody recognizeable

@JohnRennie in while loop how to set j=n?

John Rennie

j = n;
while (j > 9)
  j = j/10;

Abcd

@JohnRennie Ok I will ask later today or later sometime coz I dont understand abstract concept like streams.

@JohnRennie Which variables should be included in a variable description table?

Don't be a x-triple dot

@Zerix First, let $\vec{v_r}$ – the velocity of the ball relative to the sphere, $\vec{v}$ – the absolute velocity of the ball, $\vec{u}$ – the velocity of the semi-sphere. Detachment condition:$$N=0\implies v_r=\sqrt{gR\cos\theta}$$Break the velocities into components:$$v^2=v_x^2+v_y^2=v_{ry}^2+(u-v_{rx})^2\tag{(1)}$$Momentum conservation on Ox gives: $$Mu=mv_x,\:\:\:M=m\implies u=v_x=v_{rx}-u\implies 2u=v_{rx}$$

Going back into $(1)$:$$v^2=gR\cos\theta\sin^2\theta+\dfrac{gR\cos^3\theta}{4}=$$$$=gR\dfrac{4\c‌os\theta-3\cos^3\theta}{4}$$Conservation of energy:$$mgR(1-\cos\theta)=\dfrac{1}{2}m\left(\dfrac{v_{rx}}{2}\right)^2+\dfrac{‌1}{2}mv^2$$ Skipping algebra, this gives: $$\cos^3\theta-6\cos\theta+4=0\implies \boxed{\theta=42.94^\circ}$$

Conversation ended Feb 8, 2019 at 6:24.

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