umep physics thread

[/0]

I should have started this a long time ago... but I've been to preoccupied with my other subjects lol...

A cube of mass slides without friction at speed . It undergoes a perfectly elastic collision with the bottom tip of a rod of length and mass . The rod is pivoted about a frictionless axle through its centre, and initially it hangs straight down and is at rest. What is the cube's velocity - both speed and direction - after the collision?

(This is Question 93 from Chapter 12 of Physics for Scientists and Engineers 2nd edition - Knight)

How can I approach this question? thx

[kamil9876]

It's like the classical problem in motion in a straight line: If you have two masses undergoing an elastic collision, find a relationship between the velocities. And you do that by playing around with two simultaenous equations that you get from conservation of momentum and energy.

In this case:

Initial Kinetic Energy of cube=Final kinetic energy of cube + Final Rotational energy of rod. (1)
Initial Momentum of cube=Final Momentum of Cube + Final Momentum of Rod (2)

Equation (2) is a bit funny, because how can you equate linear and angular momentum?


The easiest way is to consider the collision that happens just as the cube is at the rod. It's path is tangential to the rod and so it's kind of like it is rotating with radius d/2, so has angular momentum of , where . The initial angular momentum of the rod is 0. The Final angular momentum of cube is and the final angular momentum of the cube is .

The thing I have in bold I don't like, but Knight actually uses it if you look at example12.23 on page 374. (In fact this info helps you with question 94 a bit more, but it will help u on this one too).

If you don't like the thing in bold just like I don't myself, I can provide an alternative using Newton's THird law which I find more rigorous (but still gives the same answer but only takes up more time).

Edit: Okay so i did the question now: Once you get those equations set up you cancel out the m and d in the momentum equation. Cancel out the m in the energy equation. The rest of the algebra should be handeled well.

2nd Edit: d/2 is radius

[/0]

Thanks kamil, but with that approach I get instead of as the answers give.

Here's my work... I've double checked the arithmetic, but maybe I made a theoretical error?



With conservation of ,







......[1]

With conservation of ,







......[2]

I then solved for in both expression because it's useless and equated them:

[/0]

Does it require more energy to escape the solar system from one point or the other in the diagram? I reckon there's a difference but how would you calculate the escape velocity if initial direction is perpendicular to gravity?


A block on a frictionless table is connected to one wall via two springs in series of spring constant and . How do you find the frequency of oscillation and the effective spring constant?


Also the following question is just for fun:
If the earth suddenly stopped its circular motion around the sun, how long would it take for it to plummet into the surface of the sun?








Thanks!

[evaporade]

(1) Yes, the solar system is not symmetrical. If the solar system is spherical, the same.
(2) 1/k = 1/k1 + 1/k2, f = 1/2pi sqrt[k1k2/m(k1 + k2)]
(3) t = integ (from Res+Rs to Rs) of 1/sqrt[2GMs(1/(Res+Rs) - 1/r)]dr

[kamil9876]

hey /0, i read your post late last night and did it twice. Once i got 5/7 first time the other time i got 1/5. The correction i made was that the formula for the rotational inertia has L not R. Hence d not d/2

[/0]

aaah I see, thx heaps kamil and evaporade

[/0]

If a hollow cylinder rolls down a frictionless hill, what is the ratio of its final centre-of-mass velocity compared to a hollow cylinder rolling down a hill with friction?
Both cylinders have mass , radius , and the hills are both height .

I don't see what the difference is mathematically... They both have the same torque so in both cases?

[dcc]

If a hollow cylinder rolls down a frictionless hill, what is the ratio of its final centre-of-mass velocity compared to a hollow cylinder rolling down a hill with friction?
Both cylinders have mass , radius , and the hills are both height .

I don't see what the difference is mathematically... They both have the same torque so in both cases?

If there is no friction, what is providing the torque?

[/0]

gravity?

[kamil9876]

Can an object roll down a frictionless surface? Think about the contact point and what happens there in rolling motion.

[evaporade]

yes, it can appear to be rolling if it is already spinning.

[kamil9876]

I'm talking about the specific case of placing a cylinder on a frictionless incline, keeping it stationary, and letting it go.

[/0]

Oh ok... I guess this is just a conceptual thing I will need to play around with.
I think it's funny how when you actually analyse the torques and hence acceleration of the cylinder you normally take the pivot to be the contact point of the cylinder, so friction (which acts from the pivot) can be ignored and gravity (acting at distance R) is the one to focus on... and yet, friction is necessary for the gravity torque to do its job!

Hmm and if the cylinder on a frictional ramp is already rolling, then since there are no external torques it must keep rolling according to conservation of angular momentum, but its angular velocity stays constant... that is very interesting.

thanks kamil and evaporade

[kamil9876]

dcc's way is also good: You can take the pivot as going through the centre of the cylinder and so the Gravity torque is 0, the Normal Torque is zero (it's parralel to radius vector(opposite direction)) but frictional torque is the only one. Hence friction is neccesary.

Wasting a lot of my time thinking about this, I think it's easier to just take the centre as pivot rather than checking if the contact point is really instantaenously stationary. The latter approach is probably easier to imagine in polygonal objects rolling down (a circle is a kind of polygon though lol)