Torque / Pulley / Weight question

[pjb73]

Hi

I'm 40 and teaching myself physics from the internet and have the following question but don't where to start!

Here is the question:
A solid disc pulley with a 5kg mass, radius of 0.5m has one layer of thin rope coiled around it. On the other end of the rope is a metal block with a mass of 10kg.
Calculate the following (without take rope friction into account. also the rope does not slip):
1. Acceleration of the falling block.
2. Velocity of the falling block at 2 metres.
3. The on the pulleys shaft (Radius of pulleys shaft is 30mm).

So far I have I drawn a diagram showing the forces on the block (F upwards and m2g down), and the forces on the pulley (F downwards at the edge of the pulley) (for both the block and the pulley, F is the force due to the rope).

Have I missed anything and what should do next?

Thanks.

 

[Aero_UoP]

Vector Mechanics for Engineers; Statics, by Beer, Johnston and Eisenberg has many examples with pulleys and ropes ;)

 

[pjb73]

Can anyone give me the link to any useful resources on the internet?

 

[Aero_UoP]

If you google the title I gave you, you'll find some pretty interesting results ;)

 

[pjb73]

Thanks Aero_UoP

 

[voko]

Have I missed anything and what should do next?

You have missed the accelerations of the block and the pulley, and accounting for them is what you should be doing next.

 

[pjb73]

Thanks for your reply voko.

ok, so I have followed the above advise. Would you please check my following calculations so I can be sure I understand correctly.

To calculate angular acceleration 1st calculate pulley inertia:

pulley inertia(I) = 1/2mr2 (2 represents squared)
I = 1/2 x 5 x 0.25
I = 0.625

acceleration(a) = hanging mass weight divided by mass of hanging mass plus pulleys inertia divided by radius squared.
a = 98 / (10 + (0.625 / 0.25))
a = 98 / (10 + 2.5)
a = 7.84 m/s

To calculate the torque:
Torque(T) = Inertia x Angular Acceleration
T = 0.625 x 7.84
T = 4.9Nm

Would someone let if the above calculations are correct?

Many thanks.

 

[voko]

acceleration(a) = hanging mass weight divided by mass of hanging mass plus pulleys inertia divided by radius squared.

The most interesting part is this equation: are you sure you can explain how you got it?

To calculate the torque:
Torque(T) = Inertia x Angular Acceleration
T = 0.625 x 7.84
T = 4.9Nm

The equation is correct; however, "7.84" is not angular acceleration; it is the (linear) acceleration of the hanging mass that you calculated above.

 

[pjb73]

Thanks for your reply voko.

I can explain how I got the acceleration equation. Is the equation correct?

Assuming the acceleration equation to be correct. To calculate the torque:
Torque(T) = Inertia x Angular Acceleration
and... Angular Acceleration = a/r = 7.84/0.5 = 15.68 m/s
T = 0.625 x 15.68
T = 9.8Nm

Is the above correct?
Also did I use the correct units for Angular Acceleration (m/s) or should it be rads/sec?
And Also, is the calculated torque on the edge of the pulley (as apposed to being on the shaft of the pulley)?

Thanks again for your help

 

[voko]

Thanks for your reply voko.

I can explain how I got the acceleration equation. Is the equation correct?

It is correct.

Assuming the acceleration equation to be correct. To calculate the torque:
Torque(T) = Inertia x Angular Acceleration
and... Angular Acceleration = a/r = 7.84/0.5 = 15.68 m/s
T = 0.625 x 15.68
T = 9.8Nm

Is the above correct?

Correct, except the units for angular acceleration.

Also did I use the correct units for Angular Acceleration (m/s) or should it be rads/sec?

Neither is correct. Linear acceleration is $m s^{-2}$, and angular acceleration is $s^{-2}$. You could say angular acceleration is $rad \cdot s^{-2}$ because radians are technically dimensionless.

And Also, is the calculated torque on the edge of the pulley (as apposed to being on the shaft of the pulley)?

Torque is the same all around. You calculated the torque assuming the force was applied at the edge.

 

[pjb73]

Thank you very much voko for all your help :)