Angular momentum Homework Problem

Click For Summary
The discussion revolves around solving angular momentum and torque problems related to a cylindrical grinding wheel and an Atwood machine. For the grinding wheel, the correct angular momentum is calculated as 6.08 kg*m^2/s, but there is confusion regarding the torque required to stop it in 6.50 seconds, with a calculated torque of 0.9347 N*m not being accepted by the system. The user expresses uncertainty about the calculations and seeks clarification on potential errors. Additionally, they present a new problem involving an Atwood machine with two masses and a pulley, asking for help in determining the acceleration of the masses and the impact of ignoring the pulley’s moment of inertia. The thread highlights common challenges in applying physics concepts to homework problems.
AstroturfHead
Messages
6
Reaction score
0
This is a problem I'm having a world of problem with:

A. What is the angular momentum of a 2.56 kg uniform cylindrical grinding wheel of radius 17.5 cm when rotating at 1480 rpm?

Correct: 6.08 kg*m^2/s

B. How much torque is required to stop it in 6.50 s?

So after 5 tries, I get that Torque = Moment of Inertia * Alpha.

Alpha = Delta (w)/ Delta (t)
Then I = 1/2*m*r^2
and I get .9347 N*m, but the computer does not. Any suggestions as to what I am doing wrong?
 
Last edited by a moderator:
Physics news on Phys.org
U've given it 4 sign.digits cf.3 before...:-p

If A is correct (manipulating units),then B is correct,as well.I don't c what it could be.

Daniel.
 
Well I have lot of these homework questions. Heres another one that I can't figure out.

An Atwood machine consists of two masses, m1 = 6.80 kg and m2 = 8.55 kg, connected by a cord that passes over a pulley free to rotate about a fixed axis

Then there's a picture but I don't know how to link to it. Its just 2 weights on the end of a string around a pully.

The pulley is a solid cylinder of radius R0 = 0.535 m and mass 0.771 kg. Determine the acceleration a of both masses. Ignore friction in the pulley bearing.

10. [1pt]
What percentage error in a would be made if the moment of inertia of the pulley were ignored? Do not enter units.

Any ideas on there either? I did the same kinda thing and said T = F*R=I*Alpha etc etc
 

Similar threads

Replies
5
Views
1K
  • · Replies 17 ·
Replies
17
Views
810
  • · Replies 9 ·
Replies
9
Views
705
Replies
26
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
18
Views
7K
Replies
335
Views
15K
Replies
13
Views
2K
Replies
3
Views
1K
Replies
3
Views
2K