Mass suspended from rotating flywheel

[Andy86]

Homework Statement

A mass of 0.5kg is suspended from a flywheel,
The mass is released from rest and falls a distance of 0.5m in 1.5s
Mass of wheel = 3kg
Outside radius of wheel = 300mm
Radius of gyration = 212mm

Homework Equations

a- Linear acceleration of wheel
b- Angular acceleration of wheel
c- Tension in rope
d- Frictional torque

The Attempt at a Solution

[/B]
a) A = 2 x S / t^2
= 2x 0.5 / 1.5^2
= 0.444m/s

b) Inertia = mass x radius of gyration^2
= 3kg x 0.212^2
= 0.134832 KG M 2

Force = mass x acceleration
= 0.5kg x 0.444
= 0.222N

Torque = Force x Distance
= 0.222N x 0.3
= 0.0666 N

ANGULAR ACCELERATION = TORQUE / INERTIA
= 0.0666N / 0.134832
= 0.4944 Rad / s

c) Tension in rope
tension = Mg - Ma
= (0.5kg x 9.81) - (0.5 x 0.444)
= 4.683N

d) Frictional torque

Torque in rope = force x dist
= 4.683 x 0.3m
= 1.4049NM

Frictional Torque = Torque - Accel. Torque
= 1.4049 - ?
= ??

Im not too sure how to find the acceleration torque! I solved these questions (or thought I did a while ago and now I am writing up my solutions to submit them some bits don't make sense, any advice or guidance would be greatly appreciated)

Thanks
A

 

[BvU]

Hi Andy, welcome to PF !

How do you distinguish friction from inertia ?
There is nothing on friction in the problem statement.
There is no question in the problem statement.

Is the radius of gyration a given or did you round off $0.3\sqrt{1\over 2}$ ?

 

[Andy86]

Hi BvU, thanks for the warm welcome!

The Radius of gyration is given in question = 212mm or 0.212m
The last question (d) asks to calculate frictional torque (resisting motion)

 

[SteamKing]

Homework Statement

A mass of 0.5kg is suspended from a flywheel,
The mass is released from rest and falls a distance of 0.5m in 1.5s
Mass of wheel = 3kg
Outside radius of wheel = 300mm
Radius of gyration = 212mm

Homework Equations

a- Linear acceleration of wheel
b- Angular acceleration of wheel
c- Tension in rope
d- Frictional torque

The Attempt at a Solution

[/B]
a) A = 2 x S / t^2
= 2x 0.5 / 1.5^2
= 0.444m/s

These are the wrong units for acceleration.

b) Inertia = mass x radius of gyration^2
= 3kg x 0.212^2
= 0.134832 KG M 2

Force = mass x acceleration
= 0.5kg x 0.444
= 0.222N

Torque = Force x Distance
= 0.222N x 0.3
= 0.0666 N

These are the wrong units for torque. Remember, the number 0.3 has units.

ANGULAR ACCELERATION = TORQUE / INERTIA
= 0.0666N / 0.134832
= 0.4944 Rad / s

These are the wrong units for angular acceleration.

c) Tension in rope
tension = Mg - Ma
= (0.5kg x 9.81) - (0.5 x 0.444)
= 4.683N

d) Frictional torque

Torque in rope = force x dist
= 4.683 x 0.3m
= 1.4049NM

Frictional Torque = Torque - Accel. Torque
= 1.4049 - ?
= ??

Im not too sure how to find the acceleration torque! I solved these questions (or thought I did a while ago and now I am writing up my solutions to submit them some bits don't make sense, any advice or guidance would be greatly appreciated)

Thanks
A

 

[BvU]

Hi BvU, thanks for the warm welcome!

The Radius of gyration is given in question = 212mm or 0.212m
The last question (d) asks to calculate frictional torque (resisting motion)

Ok, got it.

A picture is nice, I suppose you made one ?

Acceleration of the 0.5 kg weight is due to the net force on the thing. one force is mg, the other T (tension in wire)
So reconsider the torque on the wheel/motor assembly.

 

[BvU]

Next: you don't know the inertia of wheel/motor assembly, so you'll have to find the angular acceleration in a different way.

[edit] have to run now. good luck

 

[Andy86]

Think ill retype this; sorry guys

a) A=2xS/t²
= 2x0.5/1.5²
=0.444 m/s²

b) a = acceleration / radius
= 0.444m/s² / 0.3m
= 1.481 rad/s²

c) tension = Mg - Ma
= (0.5kg x 9.81m/s²) - (0.5kg x 0.444m/s²)
= 4.683N

d) Calculate frictional torque

torque in rope = force x radius
= 4.683N x 0.3M
= 1.4049NM

acceleration torque = Mass x radius of gyration² x angular acceleration
= 3 x 0.212² x 1.481
= 0.199NM

Frictional torque = torque in rope - acceleration torque
= 1.4049 - 0.199
= 1.205NM