Analyzing Rotational Work and Energy Using Newton's Laws

[Torater]

The axle of a 12.0kg solid sphere of radius 0.500m is attached to a suspended 5.00kg weight by a massless cord that passes over a frictionless, massless pulley. When the system is released, the mass drops and the sphere rolls without slipping. Find the tension in the cord and the acceleration of the mass by applying Newton's Laws.


Ok so I think I got this much right so far:

The sum of forces in the x direction: T1-fk=m1ax
The sum of forces in the y direction: T2-m2g=m2ay

And the moment of Inertia for the solid sphere on the horizontal surface is: 2/5mr^2

Please help I can't seem to solve this.

 

[tiny-tim]

Welcome to PF!

Hi Torater! Welcome to PF!

(have a tau: τ and an alpha: α and try using the X² and X₂ icons just above the Reply box )

The sum of forces in the x direction: T1-fk=m1ax
The sum of forces in the y direction: T2-m2g=m2ay

And the moment of Inertia for the solid sphere on the horizontal surface is: 2/5mr^2

The pulley is massless and frictionless, so T₁ = T₂.

You need to do τ = Iα (the rotational equivalent of F = ma) for the sphere, and use the rolling condition, a = rα.

 

[Torater]

Ok so the sum of torque equation only apply's in the horizontal direction correct? Because the mass that is being dropped only is affected in the y direction.

This is what I came up with but I don't think I did it right.

I set my accelerations equal because a closed system, as well as T1 and T2 because there is only one tension in the rope.

$$\sum$$ (torque)= T1-T2=I$$\alpha$$
Therefore I$$\alpha$$= 2/5m1r^2(a/r)

T1=m1a+m1$$\mu$$k*g=T2= m2g +m2a

therefore I solved for acceleration as:
a=m2g-$$\mu$$km1g/m1-m2

but my kinetic friction coeff is still unknown so I solved for that through the $$\sum$$ (torque) equation as so:

m1a + m1$$\mu$$kg - m2g - m2a = 2/5m1r^2 (a/r)
$$\mu$$k= 2/5m1r^2*$$\alpha$$ + m2 (g+a) -m1a/m1g

when i plug numbers in and isolate I come up with:

2/5(12)(.5^2)*(a/.5) + 5(9.81) + 5(a) - 12(a)/12(9.81)

Now i took this equation and plugged it into the acceleration equation and solved for a I came up with:

a= 5(9.81)-(1.2(a/.5) + 49.05 -7a/117.72) * 117.92 all divided by 7
7a=49.05-1.2(a/.5)+49.05+7a
0=98.1-1.2(a/.5)
a= 40.875m/s2
and then I took this acceleration and plugged it into T2 equation to solve fore the tension in the rope : T2+T1=T : 5(9.81+ 40.875) = 253.4N

I don't think that I did this right though as I used the same equation for substitution of both variables that I solved...

 

[tiny-tim]

Hi Torater!

Ok so the sum of torque equation only apply's in the horizontal direction correct?

Sorry, I don't understand that …

https://www.physicsforums.com/library.php?do=view_item&itemid=175" is about a point, not in a direction

$$\sum$$ (torque)= T1-T2=I$$\alpha$$

No!

Your τ = Iα is only for one body (in this case, the sphere), and in particular only for the torques on that body. T₂ has nothing to do with the sphere!

(and T₁ - T₂ = 0 anyway)

m1a + m1$$\mu$$kg - m2g - m2a = 2/5m1r^2 (a/r)

No … you've included T₁ in that, but T₁ acts on the axle, so it has no torque about the axle, does it?

(Your 2/5mr² is the https://www.physicsforums.com/library.php?do=view_item&itemid=31" about the axle … if you want to do torques about the point of contact, then you can use T₁, but you would need to calculate the moment of inertia about the point of contact.)

Try again. ​

 

[Torater]

I don't understand what the sum of forces in the x direction is then.

This is really frustrating me, could you just provide the sum of forces in the x direction equation then and I can try and understand it from there?


Or is my equation right other than that I need to replace my normal force with I alpha


And even if I do that I still don't have a kinetic friction coefficent given and I don't see how I can isolate these 2 equations for three variables?? (T a uk)

The pulley that is goes over is massless and frictionless so can't I just ignore its moment of inertia? and well everything to do with it in general or how does this factor into the problem?

 

[tiny-tim]

Hi Torater!

The pulley that is goes over is massless and frictionless so can't I just ignore its moment of inertia? and well everything to do with it in general or how does this factor into the problem?

It doesn't …

in fact you can treat the whole string as being in a straight line (with gravity on the 5.00kg weight only acting horizontally! )

You need a torque equation, an F = ma equation, and a rolling condition equation (and you don't need to use µ, just call the whole friction force F) …

that should do it! ​

btw, I'll be out for the rest of the evening

 

[chod]

I am having trouble with almost the same problem (could you tell me the difference between these two?)

A hoop shaped 6-kg wheel of radius 25cm is attached to a 3.00kg mass by a mass-less cord that passes over a frictionless, mass-less pulley. When the system is released the mass drops and the wheel rolls without slipping. Using Newtons laws find the tension in the chord and the mass's acceleration.

These are the formulas that I've come up with:

$$\sum$$ Fy=m2a => T-m2g=m2a

$$\sum$$ Fx=m1a => -f(static)+T=m1a$$\sum$$ t=Iα => TR+fR=Iα => TR+fR=1/2mr^2(a/R)Do these look correct?

 

[Torater]

t=Iα => TR+fR=Iα => TR+fR=1/2mr^2(a/R)

What is your fR??

aside from that it looks like the same question just with different moments of interia as I have a solid sphere in mine with an Inertia of 2/5mR^2, while you have I=mR^2

I'm just confused on how to piece all these equations together as to the friction I used kinetic not static not sure if this is going to make a difference...

 

[chod]

The fR is the frictional force*radius (I don't know whether or not this should be included). I used static friction because the rolling sphere only makes contact at one point along the horizontal, but yeah it shouldn't make a difference if you're using kinetic friction.

I'm having the same trouble as you, putting these equations together and applying the concepts is not going good with me right now. I'll give it a try and see what I come up with, I just hope someone comes by and tells me whether or not the equations are correct.

 

[chod]

Hmm I don't think the fR would be included with the rotating wheel because tension is pulling directly on the wheel whereas friction is not, so friction only comes into account on the x-axis.

 

[Torater]

Tiny Tim this is my latest attempt, but these numbers seem high still:

$$\Sigma$$ Fy= m2a=T-m2g

Fx= m1a=I$$\alpha$$ = T

2/5m1r^2(a/r) = m2(a-g)

which when numbers are put equals:

2/5(12)(.5)^2(a/r) = 5(a-9.81)

1.2[a/r] -49.05= 5a
1.2(a/0.5) -5a = 49.05

a= -18.87m/s^2


and then plugging this number back into the equation for tension is:

T=m2(a-g)
T= 5(18.87 -9.81)
T=45.3N

Does this seem right just to assume it rolls ideally without friction... otherwise with friction I don't understand how to do it.

 

[Torater]

chod: I am pretty sure friction doesn't come in on the y direction only the x direction, I just don't know how to use it properly in the x direction... If we are going to assume the pulley is frictionless, I am going to assume that the sphere rolls without friction but I am pretty sure this is a BAD assumption... I just don't understand it otherwise.

Maybe if you come up with a solution we can piece ours together and make the right one... :D

 

[tiny-tim]

welcome to pf!

Hi Torater! hi chod! welcome to pf!

(just got up :zzz: …)

A hoop shaped 6-kg wheel
…
$$\sum$$ Fy=m2a => T-m2g=m2a

$$\sum$$ Fx=m1a => -f(static)+T=m1a


$$\sum$$ t=Iα => TR+fR=Iα => TR+fR=1/2mr^2(a/R)

(have a tau: τ and an alpha: α and a sigma: ∑ and try using the X² and X₂ icons just above the Reply box )

your first https://www.physicsforums.com/library.php?do=view_item&itemid=26" equation should be m₂g - T = m₂a

and T goes through the axle, so it has no torque

and a hoop has all its mass on the rim, so as Torater pointed out, I = mr²

… it shouldn't make a difference if you're using kinetic friction.

that's right … just call the friction force F, and since it'll cancel out in the end, there's no need to say whether it's kinetic or static

Tiny Tim this is my latest attempt, but these numbers seem high still:

$$\Sigma$$ Fy= m2a=T-m2g

Fx= m1a=I$$\alpha$$ = T

2/5m1r^2(a/r) = m2(a-g)

(how can you write equations like that, with two equals signs? :yuck: if you must start with "∑F", use chod's "=>" to separate the two equations )

I'm rather confused … you seem to have missed out one of the equations (which was in your first post), and I can't work out how you eliminated F

you should be finding that the sphere effectively has has a "rolling mass" in addition to its own mass

try again in the morning! ​

 

[Torater]

∑fy=m2a = T-m2g
∑fx=m1a = T-F
∑τ= Iα = Fr sin 90 (in other-words F* lever arm)

T= m2(a+g)
T-m1a= F

2/5m1r²(a/r) = (m2(a+g) - m1a) (.5)

2/5 (12)(.5)² (a/.5) =(5(a+9.81) - 12(a)) (.5)
2.4a = (-7a + 49.05) (.5)
5.9a=24.525

a= 4.16m/s²

T= (5)(4.16+ 9.81)
T= 69.85N

 

[Doc Al]

∑fy=m2a = T-m2g

You're making a sign error here.

∑fx=m1a = T-F

OK.

When the acceleration of m1 is positive, the acceleration of m2 is down.

 

[Torater]

∑fy=-m2a = T-m2g
∑fx=m1a = T-F
∑τ= Iα = Fr sin 90 (in other-words F* lever arm)

T= m2(g-a)
T-m1a= F

2/5m1r²(a/r) = (m2(g-a) - m1a) (.5)

2/5 (12)(.5)² (a/.5) =(5(a+9.81) - 12(a)) (.5)
2.4a = (-17a + 49.05) (.5)
10.9a=24.525

a= 2.25m/s²

T= (5)(2.25+ 9.81)
T= 60.3N


Is this right now?

 

[chod]

Hi Torater! hi chod! welcome to pf!

(just got up :zzz: …)(have a tau: τ and an alpha: α and a sigma: ∑ and try using the X² and X₂ icons just above the Reply box )

your first https://www.physicsforums.com/library.php?do=view_item&itemid=26" equation should be m₂g - T = m₂a

and T goes through the axle, so it has no torque

and a hoop has all its mass on the rim, so as Torater pointed out, I = mr² that's right … just call the friction force F, and since it'll cancel out in the end, there's no need to say whether it's kinetic or static

From what you've said I got these equations:

∑Fy=m2a => T-m₂g = -m₂a (wouldn't tension for the hanging mass be positive, and the acceleration be negative?)

∑Fx=m1a => -F+T = m1a

∑t=Iα => FR=Iα => FR = mr²(a/R)
From here I'm having trouble putting these equations together, where is a good place to start?

 

[Torater]

you need to isolate for your unknown variables and chose one of the equations to plug them into to solve for the other 2 unknowns.

isolate a T and F

then try and solve

 

[chod]

you need to isolate for your unknown variables and chose one of the equations to plug them into to solve for the other 2 unknowns.

isolate a T and F

then try and solve

Alright I isolated T and F:

(1) T= -m₂(a+g)
(2) F=T-m1a

and I have FR=mR^2 (a/r) leftover

When I substitute either equation 1 or 2 into an equation I still have acceleration left over which is an unknown

 

[Torater]

No that's good if all is right ( which I am still waiting for someone to tell me I did it right haha )But with that you would have 1 unknown a

which you can substitute (1) and (2) into your FR equation and solve for a. Then use your solved a value to determine tension.I think it would work out like this based on what your equations look like:

-m2(a+g) - (m1a)(r) = m1r^2

then solve for a with your numbers of m2 m1 and r
once you have done this plug your calculated value of a back into your T equation and you have T.

 

[chod]

Yeah yours looks correct, I'm just confused as to how you got this:

2/5m1r²(a/r) = (m2(g-a) - m1a) (.5)

Can you explain how you got that please (negating the fact that the moment of inertia will be different)

 

[Torater]

2/5m1r²(a/r) = (m2(g-a) - m1a) (.5)

Okay I took my sum of Torque equation and substituted equations 1 and 2 so that I only have one unknown variable :a
I did this like so:

∑τ= Iα = Fr sin 90

Iα= (2/5m1r²)(α) which alpha is = to (a/r) and set this equal to F r so I have: (2/5m1r²)(a/r) = Fr
where F I can isolate for by this equation ∑fx=m1a = T-F (F= T-m1a)
making the equation look like:
2/5m1r²)(a/r)= T-m1a (r)

which now I need to remove T by replacing it with known variables as well which I can do from the remaining equation:∑fy=-m2a = T-m2g (T= m2(g-a))

So that I come up with the final equation with only a unknown.
2/5m1r²(a/r) = (m2(g-a) - m1a) (r)

Does that make sense?

 

[Doc Al]

∑fy=-m2a = T-m2g
∑fx=m1a = T-F
∑τ= Iα = Fr sin 90 (in other-words F* lever arm)

T= m2(g-a)
T-m1a= F

Good.

a= 2.25m/s²

Good.

T= (5)(2.25+ 9.81)
T= 60.3N

Rethink that equation. (Compare with your equation above, which is correct.)

 

[Torater]

whoops I put that into the old equation so my acceleration is correction

which I then can solve for T as: 5(9.81-2.25)
T= 37.8N

 

[Doc Al]

That's it.

 

[chod]

I'll do mine now lol

m1=6kg
m2=3kg
r=0.25m

Alright so my three equations are:

T=m2(a-g)
F=T-m1a
FR=m1r^2(a/r)

Substituting:

m1r^2(a/r) = (m2(g-a))-m1a (r)

6(0.25)^2(a/0.25) = (3(9.8-a))-6(a)(0.25)
1.5a = 29.4-3a-6a (0.25)
1.5a = -2.25a + 7.35
3.75a = 7.35
a= 1.96m/s^2


then:

T= m2(a+g)
T= 3(1.96+9.8)
T= 35.3N

 

[gneill]

Wow. Seems like a lot of work to get to this point! All those torques and sums.

A little 'cheat' I use for these 'Atwood Machine' type problems where there are rotating parts that aren't massless is to convert the rotational inertia of the parts into pseudo masses so that they look like just another mass in the system. The pseudo mass is just the moment of inertia divided by the square of the radius.

So for example, here you've got a sphere of radius r = 0.5m with a mass m1 = 12kg. It's rotational inertia is I = (2/5)*m1*r1^2. Therefore its pseudo mass is just (2/5)*m1, or 4.80kg.

Replace the sphere by a simple mass of M = m1 + (2/5)*m1 = (7/5)*m1, or 16.8kg. Now the total moving mass in the system is Mt = M + m2, where m2 is the 5.00kg block. So in particular,

Mt = (7/5)*12.0kg + 5.00kg
Mt = 21.80kg

The force that will accelerate this total mass is provided by the action of gravity on m2 alone, the dangling block. So:

F = m2*g

and in particular, F = 5.00kg*9.8m/s^2 = 49.03N giving the net acceleration for the system:

a = F/Mt = 49.03N/21.8kg = 2.25 m/s^2

 

[tiny-tim]

hi chod!

T=m2(a-g)
…
m1r^2(a/r) = (m2(g-a))-m1a (r)

(I haven't actually checked your figures)

yes, that's the right formula (you've corrected the wrong sign in the first equation),

but it would have been a lot easier if you'd noticed that everything was a multiple of r, so you can just ignore r, instead of remembering which power of 0.25 to multiply by

(this isn't a conicidence, it happens in every question of this type)

that gives you m₁a = m₂(g-a) - m₁a

A little 'cheat' I use for these 'Atwood Machine' type problems where there are rotating parts that aren't massless is to convert the rotational inertia of the parts into pseudo masses so that they look like just another mass in the system. The pseudo mass is just the moment of inertia divided by the square of the radius.

hi gneill!

yes, that's right … you can always pretend that a stationary pulley is a moving mass of I/r², and a rolling body (pulled by its axle) is a moving mass of m + I/r²

(for a spool pulled at "inner radius" s below the axle, the rolling mass is I/r(r-s), and every mass beyond the spool is effectively multiplied by 1 - s/r, since the acceleration is multiplied by that factor)

this I/r² is so useful you'd think it would have a standard name, like "effective mass" or "psuedo mass", but I don't think it does …

does anyone know of one?

in the absence of an existing standard name, I'd suggest "rolling mass"

 

[chod]

hi chod!


(I haven't actually checked your figures)

yes, that's the right formula (you've corrected the wrong sign in the first equation),

but it would have been a lot easier if you'd noticed that everything was a multiple of r, so you can just ignore r, instead of remembering which power of 0.25 to multiply by

(this isn't a conicidence, it happens in every question of this type)

that gives you m₁a = m₂(g-a) - m₁a

What you mean to say is that I can cancel out the r's across the equation?

m1r^2(a/r) = (m2(g-a))-m1a (r)

So instead I would have:

m₁a = m₂(g-a) - m₁a

 

[tiny-tim]

yes

 

[Torater]

gneil does this work for problems with inclines as well?? I have posted a question with an incline, if this trick works for that too that is very handy!

 

[gneill]

gneil does this work for problems with inclines as well?? I have posted a question with an incline, if this trick works for that too that is very handy!

It should work fine. Give it a go!

 

[Torater]

Ok take a look at my problem that has incline in the heading Ill go try it right now with this trick!

 

[tiny-tim]

Please remember that the question (particularly if it says "use Newton's laws") may not allow you to use this short-cut … so check with your professor.

However, even if you're not allowed, it's a very quick way of checking your answer (and very useful if your arithmetic sometimes goes wrong! )

 

[Torater]

ya i'll have to do it the other way because it specifies using Newtons laws but I can confirm I didn't mess up my algebra by this handy method!