Analyzing Acceleration and Forces in a Pulley System

Click For Summary
SUMMARY

The discussion centers on calculating the acceleration of a mass (m = 1100 kg) in a pulley system, where the motor produces a tension of 1.46 * 10^-4 N in the cable. The pulley has a moment of inertia of 73.8 kg*m² and a radius of 0.712 m. The correct procedure involves applying Newton's second law (F=ma) to the mass and analyzing the torques on the pulley to relate angular acceleration to linear acceleration. The final calculated acceleration of the system is 13.08 m/s², confirming the method used to derive this value.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of torque and moment of inertia
  • Familiarity with pulley systems and tension forces
  • Basic principles of angular acceleration
NEXT STEPS
  • Study the relationship between linear and angular acceleration in pulley systems
  • Learn about calculating torque in rotational dynamics
  • Explore the effects of different mass distributions on moment of inertia
  • Investigate real-world applications of pulley systems in engineering
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding dynamics in pulley systems will benefit from this discussion.

Corky
Messages
14
Reaction score
0
A motor raises a mass (m = 1100kg), it produces a tension of 1.46 *10^-4 in the cable on the right side of the pulley. the pulley has a moment of inertia of 73.8Kg*m^2 and a radius of 0.712m. The cable rides over the pulley without slipping. Determine the acceleration of mass m.

The question come with diagram with a motor beside the mass on the group, a rope goes up from the motor - around a pulley - and then back down to attach to the mass.
 
Physics news on Phys.org
Well, give it a shot. Show your work and you'll get some help.
 
Oh right forgot about that detail. Well I caculated weight = mass * gravity for the tension on the leftn side of the pully to be (1100kg) * (9.81) and from there I am not really sure. I would expect to have to subtract the given motion of inertia from the tention on the right, and then subtract the tension on the left side from that number to get the upward force on the mass. Is that right?
 
Let's do it step by step. The picture I have is a pulley with a rope hanging over it. The mass (m) is attached to the left end of the rope; a motor is attached to the right. Correct?

What you know: The tension in the right-side rope: Tright.

You also should realize: the acceleration is the same at all points along the rope. So how does the angular acceleration of the pulley relate to the acceleration of the rope? Figure that out first.

Now consider the forces on the mass: its weight pulls down, the tension in the left side rope (Tleft) pulls up. Apply F=ma to this body.

Do something similar for the pulley. There are two forces on it: the tensions of the two sides of rope. (Note: those tensions are not equal--if they were, the pulley (and rope) would not accelerate!) Now figure out what torques those tensions give to the pulley. Then apply the torque equation to this body: Torquenet = Ix(angular acceleration).
 
All right, at this point I have subtract tention left (w = 10791N) from tention right (14600N)and got a net force of 3809N. I used the equation [I(moment of intertia)= 0.5mr^2] to calculate the mass of the disk to be 291.2kg. Then I used the net force, 3809, with the equation F=ma to find the acceleration of the system to be 13.08m/s^2. Is this the correct procedure?
 
Don't skip steps. I asked you to do three things. Start with the first thing. Then we'll go from there.

1) First answer my question about how the angular acceleration of the pulley relates to the linear acceleration of the mass.

2) Then analyze the forces on the mass.

3) Then the pulley.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
2K
Replies
8
Views
14K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
6K
Replies
4
Views
835
  • · Replies 30 ·
2
Replies
30
Views
4K