Angular Acceleration of a Pulley Supporting Two Hanging Masses

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 8K views
wingman358
Messages
18
Reaction score
1

Homework Statement



Hi all! I'm working on a dynamics homework and have hit a wall here:

d27a9810-5a81-45a4-b062-9839faa86a69-original.png


"If the frictional moment at the pivot O is 2 N m, determine the angular acceleration of the grooved drum, which has a mass of 8 kg and a radius of gyration k = 225 mm.

Ans: α = 0.622 rad/s^2 "

Homework Equations



M = m g r

ƩM = I α

I = m k^2

The Attempt at a Solution



The two masses exert moments about O, so my first step is to sum the moments about the pivot O:

ƩM = (12 kg)(9.81 m/s^2)(0.2 m) - (7 kg)(9.81 m/s^2)(0.3 m)
ƩM = 2.943 N m

These moments are resisted by the frictional moment such that:

ƩM = (2.943 N m) - (2 N m)
ƩM = 0.943 N m

Next we need the moment of inertia of the pulley:

I = m_pulley * k^2
I = (8 kg)(0.225 m)^2
I = 0.405 kg m^2

Finally we can solve for angular acceleration:

ƩM = I α
so
α = ƩM / I
α = (0.943 N m) / (0.405 kg m^2)
α = 2.33 1 / s^2

My analysis seems reasonable and I come up with the right units, but it is not very close to the given answer.

Am I doing something wrong?
 
Last edited:
Physics news on Phys.org
We cannot assume that there is no acceleration on the two objects(12kg and 7kg)
 
Last edited by a moderator:
wingman358 said:

Homework Statement



Hi all! I'm working on a dynamics homework and have hit a wall here:

View attachment 152095

"If the frictional moment at the pivot O is 2 N m, determine the angular acceleration of the grooved drum, which has a mass of 8 kg and a radius of gyration k = 225 mm.

Ans: α = 0.622 rad/s^2 "

Homework Equations



M = m g r

ƩM = I α

I = m k^2

The Attempt at a Solution



The two masses exert moments about O, so my first step is to sum the moments about the pivot O:

ƩM = (12 kg)(9.81 m/s^2)(0.2 m) - (7 kg)(9.81 m/s^2)(0.3 m)
ƩM = 2.943 N m

These moments are resisted by the frictional moment such that:

ƩM = (2.943 N m) - (2 N m)
ƩM = 0.943 N m

Next we need the moment of inertia of the pulley:

I = m_pulley * k^2
I = (8 kg)(0.225 m)^2
I = 0.405 kg m^2

Finally we can solve for angular acceleration:

ƩM = I α
so
α = ƩM / I
α = (0.943 N m) / (0.405 kg m^2)
α = 2.33 1 / s^2

My analysis seems reasonable and I come up with the right units, but it is not very close to the given answer.

Am I doing something wrong?

You forgot to consider the acceleration of the angles for the two masses . That is why you acceleration is too fast. the angular acceleration of the 12kg mass and 7kg mass are missing
 
Last edited by a moderator:
Bemberson said:
You forgot to consider the acceleration of the angles for the two masses . That is why you acceleration is too fast. the angular acceleration of the 12kg mass and 7kg mass are missing
This thread is 10 years old. The user who posted it is unlikely to profit from your remark, but thank you for posting.