# Homework Help: Linear acceleration, Angular acceleration, tension.

1. Feb 6, 2017

### tom_m132

1. The problem statement, all variables and given/known data
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

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

2. Relevant equations

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

3. 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

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.

2. Feb 6, 2017

### 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.

3. Feb 6, 2017

### 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?

4. Feb 6, 2017

### BvU

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 $F_{\rm net} = ma$

5. Feb 6, 2017

### tom_m132

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!

6. Feb 6, 2017

### BvU

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

7. Feb 6, 2017

### 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.

2. Relevant equations

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

3. 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

8. Feb 6, 2017

### BvU

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

9. Feb 6, 2017

### tom_m132

Excellent, and thanks again for all your help!

10. Feb 6, 2017

### BvU

You're welcome. Good luck with your studies !