Linear acceleration, Angular acceleration, tension.

[tom_m132]

Homework Statement

A mass of 0.5 kg is suspended from a flywheel as shown in FIGURE 2.
If the mass is released from rest and falls a distance of 0.5 m in 1.5 s.
Mass of wheel: 3kg
Outside rad. of gyration of wheel: 300mm
Radius of gyration: 212mm

calculate:

(a) The linear acceleration of the mass.
(b) The angular acceleration of the wheel.
(c) The tension in the rope.
(d) The frictional torque, resisting motion

Homework Equations

a = u t + 2s / t^2
alpha = a / r

The Attempt at a Solution

a. a = u t + 2s / t^2
= 0.444 m s^-2

b. alpha = a r
= 0.444 / 0.3
= 1.48 rad s^-2

c. Now I have found online that this is T = m (g - a) However neither my learning materials or the online answers I have found offer an explanation of this, other than it being derived from the second law F = m a. Any explanation on this would be appreciated, as I am not willing to use an answer I do not understand.

d. I am yet to try this one so am not yet looking for help, no doubt I will need it soon.

I am aware these have been answered elsewhere before, but as I said I am looking to understand c. specifically.[/B]

 

[BvU]

Hello Tom,

I wonder what u is in your step a) and in your first relevant equation. Must be something from SUVAT and must be zero; am I right ?
Can't find fault with your a) and b) answers, so if we aren't both wrong, all is OK there

For c) the conceptual thing to do is draw a free body diagram for the 0.5 kg mass. Can you post it ?
The complementary FBD is for the rope revving up the wheel. That'll come in handy for part d), too.

 

[tom_m132]

Hello and thank-you!

The U in my step is is the initial velocity value, which in this case is 0 so becomes irrelevant.

I have this picture from the question:

I understand how the acceleration of the mass due to gravity acts upon the 0.5kg mass and causes the F = m g to act upon the rope, however its the subtraction of the mass times the acceleration I do not understand, I thought if anything that would also be causing tension on the rope rather than negating it as the acceleration is in the same x direction!

I'm sure there is something glaringly obvious I am missing!

Edit: I am not sure if it is to do with the mass of the flywheel oposing the freefall of the object perhaps?

 

[BvU]

causes the F = m g to act upon the rope

Not correct. $mg$ acts on the mass, not on the rope.

THE free body diagram I am waiting for () shows two forces acting on the 0.5 kg mass: the force of gravity and the tension in the rope.
The equation of motion for this mass is simply $F_{\rm net} = ma$

 

[tom_m132]

Not correct. $mg$ acts on the mass, not on the rope.

THE free body diagram I am waiting for () shows two forces acting on the 0.5 kg mass: the force of gravity and the tension in the rope.
The equation of motion for this mass is simply $F_{\rm net} = ma$

Ohh!
Gravity is acting on the mass pulling it downwards, however as we calculated it is only accelerating at 0.44 m s ^-2 not 9.81 m s ^-2 therefore the rope is subject to the tensional force of the difference between the two once plugged into the second law?
I have drawn a FBD (which is how I arrived at this conclusion) Not sure if it's quite right but it worked (assuming I am on the right track!)

Thank-you so much for the help so far (again assuming I am thinking along the right lines) it seems very obvious now!

 

[BvU]

Ohh!

I really like this kind of positive feedback: yes, a free-body diagram can really help establish insight into what's happening.

therefore the rope is subject to the tensional force of the difference between the two once plugged into the second law

If the rope weren't there, the acceleration would be $g$ indeed, so the tension in the rope must be responsible for the difference.
As you see from writing out $\ F_{\rm net} = ma \$ : $$T - mg = ma $$ (up = positive, $g$ = 9.81 m/s²) leading to $a = -9.81$ m/s² when T = 0 (as 'desired') and to $a = - 0.444$ m/s² with the value of T the exercise wants as answer to c).

 

[tom_m132]

Thank-you so much, I haven't done anything educational like this for a good few years now and am doing a distance learning course. Its very rewarding and I am enjoying it, however the lack of a lecturer is often a restriction as discussion is a brilliant way to aid understanding!

So I have had a go at part d. now that I have a figure for the tension.

Homework Equations

I = m k^2
T = I alpha
total torque = (m g - m a)r

The Attempt at a Solution

I = m k^2
= 3 x 0.212^2
= 0.1348

The torque needed to accelerate the flywheel:
t = I x alpha
= 0.1348 x 1.488
= 0.2001 N

So frictional torque is the total tourque minus the torque needed for acceleration:
Total= (m g - m a) x r
= 4.683 x 0.3
=1.405 N

Frictional torque = 1.405 - 0.2001
= 1.2049 N

[/B]

 

[BvU]

Looks good to me. One thing though: torque dimension is N m, not N

 

[tom_m132]

Looks good to me. One thing though: torque dimension is N m, not N

Excellent, and thanks again for all your help!

 

[BvU]

You're welcome. Good luck with your studies !