# Homework Help: Rotational Motion Problem - 16

1. Dec 16, 2013

### coldblood

Hi friends,
Please help me in solving this problem, I'll appreciate the help.

The problem is as:

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1493276_1461728287387610_1599334130_n.jpg

Attempt -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-ash4/q87/s720x720/1482774_1461728540720918_889630073_n.jpg

Thank you all in advance.

2. Dec 17, 2013

### tiny-tim

hi coldblood!

this is a collision

energy will be lost

both v and ω will change suddenly

hint: can you use conservation of momentum in any direction? if not, why not?

can you use conservation of angular momentum about any point? if not, why not?​

3. Dec 18, 2013

### coldblood

I think I can conserve momentum in the direction of line OP, because in this direction the wheel exerts force on the step during collision.

And the wheel rotates about point P, hence I can conserve angular momentum about that point.

4. Dec 18, 2013

### tiny-tim

hi coldblood!
no, because the impulse at the step is completely unknown …

it could be in any direction, so no direction is safe
yes, but that's not quite the right reasoning …

the impulse at the step (obviously) has no torque about the step, so there wil be no impulsive (ie sudden) change in angular momentum (the weight does have a torque, and so does the reaction force from the ground, but they're not impulsive, so they don't contribute to the impulsive change)

ok, so use impulsive conservation of angular momentum …

what do you get?

5. Dec 18, 2013

### coldblood

The torque of mg :

mg (R cos 300) But it will be anticlockwise.

The torque of reactionary force will be zero.

Last edited: Dec 18, 2013
6. Dec 18, 2013

### tiny-tim

yes, but it's not impulsive

when you do impulsive conservation of angular momentum,

you are comparing angular momentum "before" with angular momentum "after",

∆(angular momentum) = angular impulse = torque times ∆t​

and the gap, ∆t, in time between "before" and "after" is (taken to be) infinitesimally small

in that infinitesimally small time, the torque-times-time of mg is obviously infinitesimally small, ie we ignore it

(same with the reaction force: it too is not impulsive, its impulsive torque is taken to be zero)

so what impulsive forces are there?

and what is the total impulsive torque?​

7. Dec 18, 2013

### haruspex

This thread is heading in the right direction to deduce A, but I can't agree that B and C are true.

8. Dec 18, 2013

### tiny-tim

for now, let's keep our eyes firmly on A

i haven't even looked at B or C yet

9. Dec 19, 2013

### coldblood

Will the torque due to friction act as impulsive torque here or the torque due to reaction force?

Last edited: Dec 19, 2013
10. Dec 19, 2013

### tiny-tim

loosely speaking:

if the surface changes suddenly, the reaction force is impulsive; if it changes smoothly, it's not impulsive​

the step is a sudden change, so the force from it is impulsive

the ground is flat, so the force from it is not impulsive

11. Dec 19, 2013

### haruspex

Yes, that's a useful guide, but fwiw, here's how I think of it. If a force is limited in magnitude then its integral over an arbitrarily short time is arbitrarily small, so no impulsive momentum change. The gravitational force is so limited, and in consequence so are the normal force from the ground and any friction on the ground.
The normal force from the step has to deliver a certain momentum change regardless of how short the duration of impact, so the force is unlimited. Consequently, so is the frictional force.

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