I remember doing the "boring version" of this problem: the girl was replaced by a block, and the igloo was just a frictionless hemisphere with radius, R.
Could I know about the differnce with mgcos(theta) and Normal Force. I think Normal Force is triggerd by the Force(mgcos(theta)). because i am a student, so there may be something that I don't know. So I want to know what I am missing. I'm gonna wait for your answer. Sorry for my inexperience of english, because of I don't use English in my daily life.
As the rod falls down in the early part there is the force of gravity and a force exerted on the rod from the corner (wall and floor) where the rod is pushing against. That net force is not in the vertical direction but the component that the floor contributes to that net force must be vertical as there is no friction.
the friction force is opposite the velocity, v, and has a magnitude vK (v is the speed, K is the kinetic friction coefficient). Set up your eqs and try to solve! good luck!
One would expect the angle to increase. The girl starts at the top with some potential energy. The velocity, or kinetic energy, to leave the surface is the same, but the girl has to slide farther to attain the required kinetic energy and compensate for work done by the friction as she is sliding.
I looked for this for a first year univ student and was not able to find a solution at that level. The material I found suggested numerical methods or "advanced" maths was needed ref: www.researchgate.net/publication/253912967_Sliding_on_the_surface_of_a_rough_sphere
when you write equation 1, on the right side you are writing centripetal force... but you told centripetal acceleration. Now, if F=ma than is clear that is mv^2/R so it is a force but I'm a bit confused now: When do we use the force instead only the acceleration? For example, in circular motion we always use centripetal acceleration and not centripetal force to study it. Is it because in this problem we have to balance forces (and so this would be the reason why you write centripetal force instead of only the acceleration)?
The KE is 0.5*I*omega^2. omega is dphi/dt. If you can solve for phi(t), then you know what dphi/dt is at any moment in time. The moment of inertia of the rod is (1/3)M*L^2
Let suppose there is friction present with the coefficient of friction k and we have to find the angle at which the girl slipped. then will we write: - mgcos(thetha) - N = m(v^2/R) mgsin(thetha) = kN ? or we will write: - mgcos(thetha) - N = 0 mgsin(thetha) = kN ?
After watching around 10 lectures and 5 help sessions of yours, I feel that Indian teachers and Professors are best at explaining physics in a better way and in great detail. Your lectures are a helping hand to the Indian students cuz most of the problems are explained by our teachers here in India lucidly in detail.
mgl/2sinphi=1/3ml^2omega since gravity gives the torque and omega is dphi/dt so substituting that and integrating phi(t)can be obtained is this approach right?
Ok this makes sense... But how do we figure out how much time passes before this happens.. I mean, we know what the angle is when the child loses contact, but we don't know 'when' that occurs.
Centripetal force is a net force. Since the sum of all the forces in the diagram (including the centripetal force) must be 0, the centripetal force would have to be the difference between mgcos(theta) and the normal force. If mgcos(theta) were always equal to the normal force, the girl would stick to the igloo while sliding and never fall off. That doesnt hold true, though, meaning that the force of the igloo against the girl must decrease as shes sliding down.
@@joshspektor4389 Centripetal force doesn't get added up when determining the net force, it *is* a component of the net force. It is the component of the net force that acts radially inward. There is a tangential force from gravity accelerating her along the surface. We are interested in the point where the normal force becomes zero.
I'm pretty sure I would need familiarity with other more topics to solve the ruler problem so I don't think I can attempt is as I do not know much about moment of inertia or centre of mass
This is not how I would have solved the problem.. but it's a really great explanaition.. didn't understand the ruler problem though.. I kinda looks to me like it's moving forward after the collision
I think its igloo(ice) so it slides down automatically And the other reason is that we assume no velocity is provided for the body to slide down at initial point i.e topmost point
Amazing video sir. In the beginning I was unable to understand it but now when I put some effort let me try to explain this in another way. When the object slides down it slides down with zero velocity but as the velocity increases so does the amount of centripetal force required to sustain the circular motion. Here centripetal force = mgcostheta - N As the angle increases mgcostheta decreases in value but we need more centripetal force as angle increases because of simultaneous increase in velocity so N must decrease continuously Also N = mgcostheta - centripetal force When the N becomes 0 the object loses contact with the igloo. A point finally comes
Sir, the problem you begin at around 8:23 has given me a sleepless week so far. I think I'm overlooking something easy. I have the correct expression for omega, but I keep running into dead ends from there. Thanks in advance! :)
I don't really understand the addition of natural force in here. Isnt the natural force just the repulsive force? If you were standing on the ground, N would be equal to mg so I dont understand how the values of mgcos and N can be non-identical, isnt that the definition of N? How can you push on the ground harder than it pushes on you? The igloo being solid is not the reason the girl flies of, gravity is right? To me it makes more sense that the girl would fly off when the centripetal component of gravity became lesser than the tangential, which would happen at 45°.
I remember doing the "boring version" of this problem: the girl was replaced by a block, and the igloo was just a frictionless hemisphere with radius, R.
in my case friction was also present
We had the same question in our finals exam ahahahaha. And I see this video after I failed it...
Nice video. Thank you professor!
:)
Why h=R(1-cosθ ) ?????
Could I know about the differnce with mgcos(theta) and Normal Force. I think Normal Force is triggerd by the Force(mgcos(theta)). because i am a student, so there may be something that I don't know. So I want to know what I am missing. I'm gonna wait for your answer. Sorry for my inexperience of english, because of I don't use English in my daily life.
this problem is a classic - I suggest you watch the video again.
When the ruler slide their is a smile on my face😊
if i am not wrong the object still has a certain height (3:40) so how is the entire potential energy converted to kinetic energy.?????
While expressing potential energy or to be more exact while "comparing" we set a reference line which in this case we put that ball height in the end
Is the normal force from the floor to the rod in vertical direction ? and does it make a centripetal acceleration?
how many minutes into the video?
10:44
As the rod falls down in the early part there is the force of gravity and a force exerted on the rod from the corner (wall and floor) where the rod is pushing against. That net force is not in the vertical direction but the component that the floor contributes to that net force must be vertical as there is no friction.
So what makes centripetal force to the rod ? And thanks professor.
make a free-body diagram as the rod is still in contact with the wall.
Sir can u plz do a video on the same igloo problem, when friction act on the child.
the friction force is opposite the velocity, v, and has a magnitude vK (v is the speed, K is the kinetic friction coefficient). Set up your eqs and try to solve! good luck!
One would expect the angle to increase.
The girl starts at the top with some potential energy.
The velocity, or kinetic energy, to leave the surface is the same, but the girl has to slide farther to attain the required kinetic energy and compensate for work done by the friction as she is sliding.
I looked for this for a first year univ student and was not able to find a solution at that level. The material I found suggested numerical methods or "advanced" maths was needed ref: www.researchgate.net/publication/253912967_Sliding_on_the_surface_of_a_rough_sphere
when you write equation 1, on the right side you are writing centripetal force... but you told centripetal acceleration. Now, if F=ma than is clear that is mv^2/R so it is a force but I'm a bit confused now:
When do we use the force instead only the acceleration? For example, in circular motion we always use centripetal acceleration and not centripetal force to study it. Is it because in this problem we have to balance forces (and so this would be the reason why you write centripetal force instead of only the acceleration)?
v^2/R is the centripetal acceleration. mV^2/R = m*omega^2*R is the centripetal force.
Thank you sir,wish i had a teacher like you😑
Does the ruler have a (1/2Mv^2) term in the energy expression?
question unclear
At 3:07 - where does the R(1- cos(thetre)) in h = R(1 - cos(thetre)) come from?
simple trigonometry, ask your math teacher
It is equal to the whole radius R minus the part that is equal to R* cos theta, then factor R out and you get R(1- cos theta)
Is the energy when the rod is straight mgl/2 which is then mgl/2(1-cosθ)+ mv^2/2+1/2Momemt of inertia ω^2 in the question you discussed
how many minutes into the video?
The KE is 0.5*I*omega^2. omega is dphi/dt. If you can solve for phi(t), then you know what dphi/dt is at any moment in time. The moment of inertia of the rod is (1/3)M*L^2
Let suppose there is friction present with the coefficient of friction k
and we have to find the angle at which the girl slipped.
then will we write: -
mgcos(thetha) - N = m(v^2/R)
mgsin(thetha) = kN
?
or we will write: -
mgcos(thetha) - N = 0
mgsin(thetha) = kN ?
we need a diagram or FBD to clarify your doubt. Just make a proper fbd and carefully resolve the forces
After watching around 10 lectures and 5 help sessions of yours, I feel that Indian teachers and Professors are best at explaining physics in a better way and in great detail. Your lectures are a helping hand to the Indian students cuz most of the problems are explained by our teachers here in India lucidly in detail.
Just try to appreciate the effort man
Really? I have seen lots of videos on UA-cam from instructors at IIT and they are generally terrible.
what happens when there is friction is a difficult case
Hi sir, where can I see a solution to the ruler problem?
mgl/2sinphi=1/3ml^2omega since gravity gives the torque and omega is dphi/dt so substituting that and integrating phi(t)can be obtained is this approach right?
Rotational KE(t) would be 0.5*I(omega_t^2)
but how do find out that the angle at which the rod starts translating you mentioned in the video it was 48 degree ?
It's a classic physics problem. It's not a one liner. Good luck!
Ok this makes sense... But how do we figure out how much time passes before this happens.. I mean, we know what the angle is when the child loses contact, but we don't know 'when' that occurs.
What is this N stands for??
i am little bit confused...?? i think there should be mgcos(theta)=normal reaction?????????
I cannot add anything to the clarity of my correct solution.
Centripetal force is a net force. Since the sum of all the forces in the diagram (including the centripetal force) must be 0, the centripetal force would have to be the difference between mgcos(theta) and the normal force.
If mgcos(theta) were always equal to the normal force, the girl would stick to the igloo while sliding and never fall off. That doesnt hold true, though, meaning that the force of the igloo against the girl must decrease as shes sliding down.
@@joshspektor4389 Centripetal force doesn't get added up when determining the net force, it *is* a component of the net force. It is the component of the net force that acts radially inward. There is a tangential force from gravity accelerating her along the surface. We are interested in the point where the normal force becomes zero.
what makes you sir so sure that when N=0 girl falls outside the hemisphere , why not inside the hemisphere
it's a metal dome
@@lecturesbywalterlewin.they9259 what if there will be only a string lifting the girl ??
Sir work done by gravity = k2-k1
V=√usquare+2gr(1-costheta)
I'm pretty sure I would need familiarity with other more topics to solve the ruler problem so I don't think I can attempt is as I do not know much about moment of inertia or centre of mass
It was asked in Olympiad
Thank you very much sir.
This is not how I would have solved the problem.. but it's a really great explanaition.. didn't understand the ruler problem though.. I kinda looks to me like it's moving forward after the collision
Watch the video in 0.25x speed. At about halfway, it literally moves away from the stopwatch.
3:45 u=0 why professor??
I think its igloo(ice) so it slides down automatically
And the other reason is that we assume no velocity is provided for the body to slide down at initial point i.e topmost point
Amazing video sir.
In the beginning I was unable to understand it but now when I put some effort let me try to explain this in another way.
When the object slides down it slides down with zero velocity but as the velocity increases so does the amount of centripetal force required to sustain the circular motion.
Here
centripetal force = mgcostheta - N
As the angle increases mgcostheta decreases in value but we need more centripetal force as angle increases because of simultaneous increase in velocity so N must decrease continuously
Also N = mgcostheta - centripetal force
When the N becomes 0 the object loses contact with the igloo.
A point finally comes
:)
Thanks a lot sir for the reply sir, made my day. :)
You have explained it wonderfully.
Sir, the problem you begin at around 8:23 has given me a sleepless week so far. I think I'm overlooking something easy. I have the correct expression for omega, but I keep running into dead ends from there. Thanks in advance! :)
I cannot add to the clarity of this video. You should watch my 8.01 lectures covering the same topics.
Thanks! Just solved it actually!
I don't really understand the addition of natural force in here. Isnt the natural force just the repulsive force? If you were standing on the ground, N would be equal to mg so I dont understand how the values of mgcos and N can be non-identical, isnt that the definition of N? How can you push on the ground harder than it pushes on you? The igloo being solid is not the reason the girl flies of, gravity is right? To me it makes more sense that the girl would fly off when the centripetal component of gravity became lesser than the tangential, which would happen at 45°.
9min 1second
Hindi
La
sir sir sir sir