Torque - Block's cord wrapped around a flywheel

[MathewsMD]

Torque -- Block's cord wrapped around a flywheel...

Attempt:

FBD: mg acts tangentially to the final point of contact b/w string and the pulley. F = mg

T = Iα = rF

αr = a = r²F/I = 1.28g

I don't see any other forces needed to be assessed, and my calculation seem correct. Any help would be greatly appreciated!

BTW:
http://answers.yahoo.com/question/index?qid=20120331140304AAqf7Zk
is a solution but why do we consider there to be a T? Doesn't taking into consideration the rotational inertia enough? I just don't quite fully understand the reasoning for why T is included...

 

[rock.freak667]

The forces acting on the 16 kg mass are its weight 'mg' and the tension in the cord T.

So as the mass falls down the resultant force on it is ma = mg-T.

 

[MathewsMD]

The forces acting on the 16 kg mass are its weight 'mg' and the tension in the cord T.

So as the mass falls down the resultant force on it is ma = mg-T.

1. Ok, so just to confirm: if there was another mass on the other side of the pulley attached to the same string, it would also have the same tension, right?

2. This gets a bit more confusing to describe, but what if there were now two strings on the pulley, and 2 masses connected to each end of the string. So there is a total of 4 objects, and they are of different masses. The tension in the two strings, assuming they were of negligible mass, would have to still be calculated separately, right?

How exactly would this look in the equation for torque:

For 1,
m is mass 1 and M is mass 2, and m is the bigger mass
mg - T = ma
T - Mg = Ma
(m-M)g/(m+M) = a
torque = I_pulleya/r = I<sub>pulley(m-M)g/(m+M) /r right?

For 2,

I'm slightly confused myself, but here's a go at it...
mg - T = ma
T - Mg = Ma
(m-M)g/(m+M) = a for one string

m2g - T = m2a
T - M2g = M2a
(m2-M2)g/(m2+M2) = a2

Just simply add the two acceleration and find torque like I did in the end of part 1. Any comments?


Also, if the masses are the same, and no external torque is applied, then there is no rotation in the pulley, right?</sub>

 

[MathewsMD]

?

Any help would be greatly appreciated!

 

[rock.freak667]

1. Ok, so just to confirm: if there was another mass on the other side of the pulley attached to the same string, it would also have the same tension, right?

Once the entire system is in equilibrium then the tendons would be equal.

2. This gets a bit more confusing to describe, but what if there were now two strings on the pulley, and 2 masses connected to each end of the string. So there is a total of 4 objects, and they are of different masses. The tension in the two strings, assuming they were of negligible mass, would have to still be calculated separately, right?

How exactly would this look in the equation for torque:

For 1,
m is mass 1 and M is mass 2, and m is the bigger mass
mg - T = ma
T - Mg = Ma
(m-M)g/(m+M) = a
torque = I_pulleya/r = I<sub>pulley(m-M)g/(m+M) /r right?

For 2,

I'm slightly confused myself, but here's a go at it...
mg - T = ma
T - Mg = Ma
(m-M)g/(m+M) = a for one string

m2g - T = m2a
T - M2g = M2a
(m2-M2)g/(m2+M2) = a2

Just simply add the two acceleration and find torque like I did in the end of part 1. Any comments?


Also, if the masses are the same, and no external torque is applied, then there is no rotation in the pulley, right?</sub>

_{If I am understanding your question properly, yes you would need to calculate the acceleration for each pulley system separately and then add up the accelerations acting on the pulley itself.}