Motion of a Uniform Stick on a Frozen Lake After an Impulse

[ZxcvbnM2000]

Homework Statement

A straight uniform stick of L= 1m and mass m = 1 KG lies on a surface of a frozen lake.Someone kicks the stick at one end,imparting an impulse of 2Ns normal to the stick.Describe the subsequent motion of the stick.

Homework Equations

I=mΔV

The Attempt at a Solution

Obviously the first thing we have to do is find the linear velocity at the end of the stick ( actually all molecules of the stick will be moving with that velocity). so I=m(V1-V0)

<=> 2=1*V1 , V0=0 since it was at rest.Therefore V1=2m/s which will remain constant due to lack of friction.The stick will be sliding forward since there are no forces to counteract with the initial force.But this force also created a torque which causes the object to rotate.

Should i assume that since there is no friction , the stick is rolling right from the beginning ? How can i find the angular velocity ?

I believe that since the kick is an external force and thus an external torque then linear and angular momentum are not conserved .

Any ideas ?

The force of the kick also caused the object to rotate

 

[ZxcvbnM2000]

Well i don't think it's rolling otherwise Vcm= wR <=> w= 2/0.5 = 4 rad/s which is not the right answer . Hmm,so we have an object that is slipping and rotating and there is no friction ... :S Grrr !

 

[ZxcvbnM2000]

No one ? :(

 

[tiny-tim]

Hi ZxcvbnM2000!

(erm … NEVER reply to your own question, it takes you off the "no replies" list!)

just use impulse = change in momentum

and torque of impulse = change in angular momentum (about the same point) …

what do you get? ​

(and I'm off to bed :zzz:)

 

[ZxcvbnM2000]

I understand that Impulse = change in momentum

but why does torque of impulse = change in angular momentum ?

Impulse is not a force it is Ft :S So how can it create a torque . Can you explain this physically ?

Let me check if what u say is making sense unit-wise .

R*mat = Iω Kg*m*m /s = Kg*m*m*rad/s which is true since rad/s = 1/s since ω is also measured in hertz ... hmm

i still don't get it though.

1) The stick is stationary . I kick its end . Will it begin rotating and slipping at the same time or not ? . I am lost :S

Assuming that i understood everything let me proceed to the calculations..

2=1*V so Vcm = 2 m/s

mVcmR = Iω Ι = 1/12 ΜL^2 so Icm = 1/12 therefore 1*2*0.5*12 = ω

So Vcm = ωR is not valid in this case because if it were true then the stick would be rolling , right ?

Damn rotational and translation is soooo confusing .

Anyway , have a good night and thanks for your time :)

 

[tiny-tim]

Hi ZxcvbnM2000!

(try using the X² and X₂ buttons just above the Reply box )

2=1*V so Vcm = 2 m/s

mVcmR = Iω Ι = 1/12 ΜL^2 so Icm = 1/12 therefore 1*2*0.5*12 = ω

yes that's all correct

(except that your mV_cmR should be the torque, which in this case of course is the same)

so it's rotating at 12 rad/s, and moving at only 2 m/s

(the question doesn't ask for it, but can you find the centre of rotation?)

So Vcm = ωR is not valid in this case because if it were true then the stick would be rolling , right ?

i don't understand what you mean by rolling … there's nothing for it to roll against

I understand that Impulse = change in momentum

but why does torque of impulse = change in angular momentum ?

Impulse is not a force it is Ft :S So how can it create a torque . Can you explain this physically ?

I dω/dt = torque = force.distance

so Iω = ∫ torque dt

= ∫ (distance x force) dt

= distance x (∫ force dt)

= distance x impulse ! ​

(eg, in this case, if you applied the same impulse slowly, eg 0.5 N for 4 s (total 2 N.s),

then for 4 s the angular acceleration would be dω/dt = torque/I = 0.5 R/I (a constant),

and so the total increase in angular velocity is ω = torque*time/I = 0.5 IR times 4 = 2 IR)

 

[ZxcvbnM2000]

Well i already said the moment of inertia is M^L/12 so i assumed that it is the centre of mass of the stick . Can i use it as a general rule and say that for anybody which is free to move any force applied to it ( besides those who cross the axis of rotation of the CoM) will cause it to rotate about its CoM.

In terms of energies : E_initial = 1/2 ( I^ω +mV^cm) .

Why isn't V_cm = ωR ? Does this means that a particle at the end of the stick will cover more distance than the CoM in the same time t ?

:)

 

[tiny-tim]

Well i already said the moment of inertia is M^L/12 so i assumed that it is the centre of mass of the stick . Can i use it as a general rule and say that for anybody which is free to move any force applied to it ( besides those who cross the axis of rotation of the CoM) will cause it to rotate about its CoM.

a body rotates about its centre of rotation

(the clue's in the name! )

the formulas torque = Iα and torque of impulse = I(∆ω) only (and always) work about the center of mass and the centre of rotation …

in this case it isn't obvious where the centre of rotation is, so we use I and torque about the centre of mass …

that doesn't mean it's rotating about the centre of mass! ​

 

[ZxcvbnM2000]

That's very useful thank you !

Lastly, can you tell me whether what i said about Vcm = ωR is true or nonsense ?

That's my last question ever...about rotation :p

 

[tiny-tim]

Why isn't V_cm = ωR ? Does this means that a particle at the end of the stick will cover more distance than the CoM in the same time t ?

the c.o.m. moves in a straight line, and the stick revolves around that moving point (so yes, the rest of the stick moves further than the c.o.m.)

if V_P is the velocity of a point P at distance R from the c.o.m., then

|V_P - V_cm = ωR …​

(in 3D: V_P - V_cm| = ω × R)

so V_cm = ωR only if P is stationary