u said if no value of coefficient friction is given we can assume pure rolling motion in pulley question,can this concept extended to objects rolling on an incline or objects moving on horizontal surface?
And yes if no value of friction, it means no friction,right ?
As a general rule if the object is sliding then the question will either tell you the friction coefficient or it will tell you the surface is frictionless.
If it doesn't say anything that usually means the friction is high enough that no sliding takes place.
If you have a particular question in mind you can post it and we can have a look at it.
@Bhavay I'd guess if a question tells you an object is rolling it normally means pure rolling unless it specifically says otherwise.
@Bhavay that means pure rolling. It doesn't specifically say the object is slipping, and if the object was slipping the queston cannot be answered because α and a would not be related.
There's one more thing i would like to say , At this point i am self preparing JEE aspirant , and my exam is in next month so i will be be asking a lot of questions sir.
Sir for next question , i got my correct answer.Although it doesn't ask for reason , i am providing u the reason( that i think is correct) and i wish u too check that what i think is actually correct or not.
If the particle and rod were on a frictionless surface then you can use conservation of linear momentum. After the collision the combined rod and particle would end up moving to right.
But with a pivot securing the rod to the surface linear momentum is not conserved. Obviously not since after the collision the linear momentum is zero since the rod just rotates about a fixed point.
@Bhavay all SHM is related to angular motion. Even with a simple pendulum you can use either sideways distance moved by the bob or angle of the string rotated from the vertical. Both give the same answer.
Yes, you end up with the torque approximately proportional to the angle of rotation. That's using the small angle aproximations cos(θ) = 1 and sin(θ) = θ
They are calculating the MOI about the point of contact with the ground. The first part is just the parallel axis theorem. The MOI of a disk about its centre is ½MR² so moving a distance R gives you the MOI about a point on the edge as ½MR² + MR². OK so far?
For a small angle we can take the red dot as approximately fixed.
You're right it wouldn't be fixed because the disk would roll sideways, but if the rotation is small enough the sideways motion is smll enough to take the red dot as fixed.
That looks straightforward. It's just a complicated way of setting the velocity of the mass on a spring i.e. when the spring extension is zero the velocity of the mass is given by ½mv^2 = mgh i.e. v² = 2gh
It says the mass of the pan is zero, so the velocity of the mass doesn't change when it hits the pan.
But if you consider the mass hanging motionless from the spring there would be an extension given by kx = mg. So the mean position is a distance x = mg/k below the point at which the mass hits the pan.
At the instant the mass attaches to the spring it has fallen a distance h so the velocity is v² =2gh, but it is still above the equilibrium point. It has to travel down a farther distance x = mg/k to reach the equilibrium point.
Yes. At the moment the mass attaches to the spring the KE of the mass is mgh. But the spring is displaced a distance x = mg/k from the equilibrium point, so the spring still has a PE ½kx² and the total energy is the sum of these two.
So the total energy is mgh + ½m²g²/k
So the amplitude of the vibration about the new equilbrium point is found by equating this to the PE:
If we do this rigorously we would say h is a negative number because the height has decreased (assuming upwards is positive) but g is also negative because g points downwards. So the product gh is positive.