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Problem 1.136 from Irodov's book

  • #1
911
18
Hello

I am attaching a problem from Irodov's book "Problems in general physics". There is also a website where the solutions are given. I am just trying to understand the solution. Solution is given at http://irodovsolutionsmechanics.blogspot.com/2008_05_03_archive.html

Now in the problem I assumed that the particle keeps in touch with the trajectory for all the half circle. I think its a wrong assumption. Also the problem doesnt mention if the body starts rolling from the rest. does it matter ? In the solution given I have not understood how he got equation 1. If we use conservation of energy , then we will need to take care of the kinetic energy at the bottom. He doesnt use that kinetic energy at all. Also the problem just asks for the velocity of the body after breaking off the groove but this guy calculates the horizontal component of the velocity. Even at the back of the book, the horizontal component of the velocity is given. can somebody clarify ?

Issac N

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
430
2
The body does break off the groove. It has to have a certain minimum velocity at the base to complete a circle. Assuming a complete circle is incorrect.
The eqn.1 does use k.e at the bottom but that is also equal to g.p.e at the top of the hill.
They also have mentioned zero velocity at the top - look what they say-"starts sliding".
And they use the horizontal velocity because they are originally asking for velocity at the highest point of trajectory.
 
  • #3
911
18
ok the equation 1 now makes sense. but why do we need to take horizontal component
of the velocity. thats the answer at the end of the book too. at the time of breaking from the groove, v = sqrt(gh/3) but the book gives v= (2/3) sqrt(gh/3)... as given in the above solution.

issac n
 
  • #4
Doc Al
Mentor
44,909
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but why do we need to take horizontal component
of the velocity.
Because at the highest point of the trajectory the vertical component will be zero.
at the time of breaking from the groove, v = sqrt(gh/3)
How did you derive that?
 
  • #5
rl.bhat
Homework Helper
4,433
7
ok the equation 1 now makes sense. but why do we need to take horizontal component
of the velocity. thats the answer at the end of the book too. at the time of breaking from the groove, v = sqrt(gh/3) but the book gives v= (2/3) sqrt(gh/3)... as given in the above solution.

issac n
In the solution of problem, they have found the velocity of the ball in the position where the line joining the ball and the center makes an angle θ with the horizontal.
In your attachment, the velocity at the highest point of the semicircle is asked.
 
  • #6
911
18
Because at the highest point of the trajectory the vertical component will be zero.

How did you derive that?
Doc Al

if you see the solution provided at the link, when the object breaks off the groove, the normal reaction N becomes zero .so plugging N=0 in eq. 2 we can solve for sin(theta). also the problem asks to find the velocity at the highest point of the trajectory of the object, not the highest point of the half-circle. please see the solution link given by me. i didn't understand the last part where he says v= (2/3) sqrt(gh/3)

Issac N
 
  • #7
Doc Al
Mentor
44,909
1,169
Doc Al

if you see the solution provided at the link, when the object breaks off the groove, the normal reaction N becomes zero .so plugging N=0 in eq. 2 we can solve for sin(theta). also the problem asks to find the velocity at the highest point of the trajectory of the object, not the highest point of the half-circle. please see the solution link given by me.
My bad. I misread what you wrote.

i didn't understand the last part where he says v= (2/3) sqrt(gh/3)
You have the velocity at the break point. Find its horizontal component. That won't change as the object reaches its highest point.

Once the object leaves the surface, it's a free projectile under gravity. For a projectile, treat horizontal and vertical motion separately. Only the vertical speed changes as the object rises; the horizontal speed is constant. At the top, the vertical component is zero.
 
Last edited:
  • #8
rl.bhat
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Will you please post the full text of the problem?
 
  • #9
Doc Al
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Will you please post the full text of the problem?
See the attachment in post #1.
 
  • #10
rl.bhat
Homework Helper
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See the attachment in post #1.
OK. Got it.
 
  • #11
911
18
My bad. I misread what you wrote.


You have the velocity at the break point. Find its horizontal component. That won't change as the object reaches its highest point.

Once the object leaves the surface, it's a free projectile under gravity. For a projectile, treat horizontal and vertical motion separately. Only the vertical speed changes as the object rises; the horizontal speed is constant. At the top, the vertical component is zero.
thanks.... makes sense now
 

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