Rotational work and energy

  • Thread starter Torater
  • Start date
  • #26
11
0
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
 
  • #27
gneill
Mentor
20,840
2,814
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
 
  • #28
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi chod! :smile:
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 :wink:

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

that gives you m1a = m2(g-a) - m1a :smile:
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! :smile:

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" o:)
 
  • #29
11
0
hi chod! :smile:


(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 :wink:

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

that gives you m1a = m2(g-a) - m1a :smile:

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
 
  • #30
tiny-tim
Science Advisor
Homework Helper
25,832
251
yes :smile:
 
  • #31
27
0
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!
 
  • #32
gneill
Mentor
20,840
2,814
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!
 
  • #33
27
0
Ok take a look at my problem that has incline in the heading Ill go try it right now with this trick!
 
  • #34
tiny-tim
Science Advisor
Homework Helper
25,832
251
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. :smile:

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! :wink:)
 
  • #35
27
0
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!
 

Related Threads on Rotational work and energy

  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
14
Views
1K
Replies
1
Views
2K
Replies
4
Views
10K
Top