Rotational Motion and velocity

[zorro]

Homework Statement

An inelastic uniform sphere of radius 'a' is sliding without rotation on a smooth horizontal plane when it impinges on a thin horizontal rod at right angles to its direction of motion and at a height 'b' from the plane. Find the maximum velocity so that it will just roll over the rod.

The Attempt at a Solution

Let the initial velocity of the sphere be v and its final angular velocity be ω

By conservation of angular momentum about the corner point,
mv(a-b) = 7/5 ma²ω

By conservation of energy,

0.5mv² = mgb + 0.5*7/5*ma²ω²

On solving, I got

which is incorrect :(
Help!

 

[tiny-tim]

Hi Abdul!

An inelastic uniform sphere of radius 'a' is sliding without rotation on a smooth horizontal plane when it impinges on a thin horizontal rod at right angles to its direction of motion and at a height 'b' from the plane. Find the maximum velocity so that it will just roll over the rod.
…
By conservation of angular momentum about the corner point,
mv(a-b) = 7/5 ma²ω

I don't understand this …

the weight will have a torque so why would angular momentum be conserved?

And if it "just" rolls over, won't ω be taken as 0?

 

[zorro]

The torque due to weight becomes an internal torque if you consider the Earth and the sphere to be your system. The net internal torques add to zero.

It just rolls over means that it just starts rolling without slipping motion (as opposed to earlier slipping motion)

 

[tiny-tim]

The torque due to weight becomes an internal torque if you consider the Earth and the sphere to be your system. The net internal torques add to zero.

You can't do that! …

you need to find the motion of the sphere relative to the Earth!

It just rolls over means that it just starts rolling without slipping motion (as opposed to earlier slipping motion)

Ah, I see what you mean …

no not in this case …

there are sometimes questions about things slipping and then starting rolling (but still moving of course), but this question is envisaging the sphere hitting the step, immediately starting to roll up onto the step, and finally juuuuust making it, at zero speed (and zero angular speed)!

 

[zorro]

You can't do that! …

you need to find the motion of the sphere relative to the Earth!

The velocity 'v' is relative to the earth.

Ah, I see what you mean …

no not in this case …

there are sometimes questions about things slipping and then starting rolling (but still moving of course), but this question is envisaging the sphere hitting the step, immediately starting to roll up onto the step, and finally juuuuust making it, at zero speed (and zero angular speed)!

eh? in that case the solution will be too simple.
v will be (2gb)^0.5, independent of 'a'.

 

[tiny-tim]

Hi Abdul!

(just got up :zzz: …)

v will be (2gb)^0.5, independent of 'a'.

I think that's right.

 

[zorro]

Sorry that is incorrect.

 

[hikaru1221]

There is nothing wrong with the angular momentum conservation equation. The wrong part lies in the energy conservation equation. Since the collision is inelastic, energy before collision < energy after collision. However, when the sphere climbs up (after collision), energy is conserved:
$$\frac{1}{2}.\frac{7}{5}ma^2\omega_0 ^2=mgb$$
where $$\omega _0$$ is the initial angular speed right after the collision.

There is also a constraint for b. Look at the result, one can tell that a>b is the condition. However, that is not enough. We should also notice that, in order to perform rotation about the rod, the sphere must have ENOUGH centripetal force. Since the force by the rod is outwards, this makes things harder:
$$F_{centripetal}=mg\frac{a-b}{a}-N$$
where N is the radial component of the force by the rod. We also have:
$$F_{centripetal} = m\omega ^2a$$
Therefore:
$$\omega ^2 a = g\frac{a-b}{a}-\frac{N}{m}$$
We must have the above equation satisfied at all time after the collision. Try to find the condition

 

[tiny-tim]

… Since the collision is inelastic…

hi hikaru1221!

thanks, i hadn't noticed that …

the step is a sudden obstacle, not a gradual slope, so energy isn't conserved at the moment of collision

sorry, Abdul!

 

[zorro]

There is also a constraint for b. Look at the result, one can tell that a>b is the condition. However, that is not enough. We should also notice that, in order to perform rotation about the rod, the sphere must have ENOUGH centripetal force. Since the force by the rod is outwards, this makes things harder:
$$F_{centripetal}=mg\frac{a-b}{a}-N$$
where N is the radial component of the force by the rod. We also have:
$$F_{centripetal} = m\omega ^2a$$
Therefore:
$$\omega ^2 a = g\frac{a-b}{a}-\frac{N}{m}$$
We must have the above equation satisfied at all time after the collision. Try to find the condition

That condition is that the minimum value of the normal force is greater than 0
b < 7a/17
Thank you very much!

thanks, i hadn't noticed that …

No problem. Even I didnot notice that

 

[hikaru1221]

There is also another condition. Notice that $$\omega$$ I wrote in that equation is not $$\omega _0$$; it is angular speed at ANY TIME after collision.

 

[zorro]

Oh so you mean ω <= ωo ?

 

[hikaru1221]

Yep.
Seems too obvious, right? For harder problems, you would never see such obvious things; at those times, you have to come back to the foundation of the condition

 

[tiny-tim]

hi hikaru1221!

(have an omega: ω )

There is also another condition. Notice that $$\omega$$ I wrote in that equation is not $$\omega _0$$; it is angular speed at ANY TIME after collision.

but ω will decrease after the collision, so the only effective constraint is on ω₀.

 

[hikaru1221]

but ω will decrease after the collision, so the only effective constraint is on ω₀.

Yes, of course it is, but just for this specific problem. I just want to remind Abdul to keep in mind the fundamental before proceeding. He's practicing for important exams, so it's better for him not to let go of any mark And should he face harder problems in the exams and the condition is not so obvious, he should also go back to the fundamental. Exam can test candidates' thinking at any point.