How many radians does the cylinder rotate in the first 5.0 seconds?

  • Thread starter Thread starter jy1231
  • Start date Start date
  • Tags Tags
    Dynamics
Click For Summary
SUMMARY

The discussion centers on calculating the angular displacement of a solid cylinder with a mass of 10 kg, pivoted about a frictionless axis, under the influence of two forces: F1 = 5.0 N and F2 = 6.0 N. The torques generated by these forces are T(1) = 5 N*m and T(2) = 3 N*m, resulting in a net torque of 2 N*m in the clockwise direction. To find the angular acceleration, the moment of inertia must be determined, followed by using the angular displacement formula to calculate the total radians rotated in the first 5.0 seconds.

PREREQUISITES
  • Understanding of torque calculations in rotational dynamics
  • Knowledge of moment of inertia for solid cylinders
  • Familiarity with angular acceleration and displacement equations
  • Basic principles of forces and their effects on rotational motion
NEXT STEPS
  • Calculate the moment of inertia for a solid cylinder using the formula I = 0.5 * m * R^2
  • Learn how to apply the angular acceleration formula α = τ/I
  • Explore the relationship between angular displacement and time using θ = ω₀t + 0.5αt²
  • Investigate the effects of multiple forces on rotational motion and net torque
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding rotational dynamics and torque calculations in mechanical systems.

jy1231
Messages
2
Reaction score
0
1. Homework Statement

A solid cylinder of mass 10 kg is pivoted about a frictionless axis through the center O. A rope wrapped around the outer radius R1 = 1.0 m, exerts a force F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force F2 = 6.0 N downward. How many radians does the cylinder rotate through in the first 5.0 seconds, if it starts from rest?

2. Homework Equations
3. The Attempt at a Solution

I did find the torque for each force, with the counterclockwise force being positive.

so

T(1) = (1.0 m)(5m) = 5 N*m
T(2) = (0.5 m)(6.0 N) = 3 N*m

so I figured out that the net torque would be 2 N*m in the clockwise direction.

But i don't know what to do with this.

Thank you so much!

Sorry, here's the picture: http://tinypic.com/view.php?pic=otlszm&s=7
 
Last edited:
Physics news on Phys.org
Find the angular acceleration from another formula for torque (which also requires the moment of inertia)
 
jy1231 said:
1. Homework Statement

A solid cylinder of mass 10 kg is pivoted about a frictionless axis through the center O. A rope wrapped around the outer radius R1 = 1.0 m, exerts a force F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force F2 = 6.0 N downward. How many radians does the cylinder rotate through in the first 5.0 seconds, if it starts from rest?

Hmm. Was there a diagram to accompany this problem? How are we to know if the ropes are wrapped in the same direction?
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 97 ·
4
Replies
97
Views
6K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
335
Views
17K