# How long will it take for one cylinder to turn one revolution?

1. Mar 5, 2012

### ScienceGeek24

1. The problem statement, all variables and given/known data

A cylinder is initially at rest, how long will it take for the cylinder to turn one revolution?

2. Relevant equations

(Theta)f=(Theta)i+(alpha)(deltaT)

3. The attempt at a solution

I know that the distance of the cylinder to turn one revolution is 2Pi over speed. My speed was 0.05 rad/s so i divided 2Pi/0.05 rad/s and it did not give me the right time which is 15.9 sec. Help?

2. Mar 5, 2012

### SammyS

Staff Emeritus
Somehow, this problem, as stated, is rather incomplete.

If the cylinder is initially at rest, what makes it move at all? ... or what describes its subsequent motion?

3. Mar 5, 2012

### ScienceGeek24

ok sorry for not being specific.

this question is the second part of the firsst question which is this one.

A solid cylinder is pivoted about a frictionless axle as shown. A rope wrapped around the outer radius of 2 m exerts a downward force of 3N. A rope wrapped around the inner radius of 0.7 m exerts a force of 8 N to the right. The moment of inertia of the cylinder is 8 kg m^2. Find the angular acceleration.

I found the solution to this one by t=+-rFsin(theta) here is the specifics https://www.physicsforums.com/showthread.php?t=583881

However, my solution to the first part of the problem was the sum of all torque forces Sumt=I(alpha)

-Rfsin(theta)+rFsin(theta)=I(alpha) I had my Moment of inertia giving my the problem so all i had to do was to solve for (alpha) which gave me 0.05 rads/s.

The second part of the problem says,

If the cylinder in problem 10 is initially at rest, how long will it take for the cylinder to turn one revolution??

4. Mar 5, 2012

### ScienceGeek24

That is the part where i use (Theta)f=(Theta)i+(alpha)(deltaT) and solve for t but it did not give me the right answer.

5. Mar 5, 2012

### SammyS

Staff Emeritus
Assuming that α is angular acceleration, it should have units of rad/s2 , not rad/s .

6. Mar 5, 2012

### ScienceGeek24

yes yes sorry for that one.

7. Mar 5, 2012

### Staff: Mentor

Angular kinematic equations have direct parallels to their linear counterparts. If θ is angular position, and $\alpha$ is angular acceleration, then your formula above is equivalent, in linear terms, to

$x_f = x_i + at$

which is not correct. Appropriate linear formulas are:

$v_f = v_i + a t$ and $x_f = x_i + v_i t + (1/2)a t^2$.

and their angular counterparts:

$\omega_f = \omega_i + \alpha t$ and $\theta_f = \theta_i + \omega_i t + (1/2)\alpha t^2$.

8. Mar 5, 2012

### ScienceGeek24

hmm which one should I use because I dont have angular velocity I only have angular acceleration. How can i start up with?

9. Mar 5, 2012

### Staff: Mentor

You drop any terms that are zero from the equations. For example, if ωi is zero, then ωit will always be zero.

10. Mar 5, 2012