What is the acceleration of the center of mass of the cylinder?

[cloudzer0]

If you guys would be kind enough to show me if my attempt to problems posted are correct or incorrect.

Homework Statement

A uniform 3.0 kg cylinder of radius 0.15 m is suspended by two strings wrapped around it, as shown in the figure below. As the cylinder descends, the strings unwind from it. What is the acceleration of the center of mass of the cylinder? (Neglect the mass of the string.)
http://www.webassign.net/wb/8-43.gif

Homework Equations

α = alpha

rmg = ½mr²
a = rα

The Attempt at a Solution

2g = rα
2g/r = α

Once I know α, I can plug into a = rα
and my answer is 19.6 m/s²
. Homework Statement
In the figure below, a constant horizontal force Fapp of magnitude 15 N is applied to a wheel of mass 9 kg and radius 0.70 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.67 m/s2.
http://www.webassign.net/hrw/hrw7_11-30.gif

a)In unit-vector notation, what is the frictional force on the wheel?
b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

Homework Equations

ma-F =Ffriction

T =Iα
α = a/r

The Attempt at a Solution

a) ma- F = Ffriction
(9kg)(0.67m/s²) - 15 N = 8.97 N

b) T =Iα
(Ffrictional)(r) = (I)(a/r)
(Ffrictional)(r²)/(a) = I

6.56kg m² = I

Homework Statement

The body in the figure is pivoted at O. Three forces act on it in the directions shown on the figure. FA = 12 N at point A, 8.0 m from O. FB = 16 N at point B, 4.0 m from O. FC = 19 N at point C, 3.0 m from O. What is the net torque about O?
http://www.webassign.net/hrw/11_39.gif

Homework Equations

T = r F sinФ

The Attempt at a Solution

TA = (8m)(12N)(sin45) = 67.88 N m
TB = (4m)(16N)(sin90) = 64 N m
TC = (3m)(19N)(sin20) = 19.50 nm

Tnet = 151.38 N m

Homework Statement

(a) How many uniform, identical textbooks of width 30.0 cm can be stacked on top of each other on a level surface without the stack falling over if each successive book is displaced 2.50 cm in width relative to the book below it?

(b) If the books are 5.00 cm thick, what will be the height of the center of mass of the stack above the level surface?

The Attempt at a Solution

There is no equation for me to solve.. I just did this problem by attempting to find the Center of Mass for the books as it stacks up. By drawing out a picture, I found that the for part a) it would take a total of 7 books.

b) If there are 7 books total, the stack center of mass would be 3.5 cm. Finally, take the center of mass(3.5cm) and multiply by the thickness(5.0cm) of each book to find the height. 17.5cm

 

[alphysicist]

Hi cloudzer0,

I just looked at the first one below:

If you guys would be kind enough to show me if my attempt to problems posted are correct or incorrect.

Homework Statement

A uniform 3.0 kg cylinder of radius 0.15 m is suspended by two strings wrapped around it, as shown in the figure below. As the cylinder descends, the strings unwind from it. What is the acceleration of the center of mass of the cylinder? (Neglect the mass of the string.)
http://www.webassign.net/wb/8-43.gif

Homework Equations

α = alpha

rmg = ½mr²

This does not look right to me; notice that the units are different on either side of the equation.

a = rα

The Attempt at a Solution

2g = rα
2g/r = α

Once I know α, I can plug into a = rα
and my answer is 19.6 m/s²

This cannot be right. Since there are no downward forces except gravity, the largest acceleration would be 9.8m/s².

I don't see what you have done to find alpha. To find the answer, draw a force diagram and write out Newton's law for the forces, and a separate equation for the torques. You should then be able to find a. What do you get?

 

[cloudzer0]

Thank you for your reply alphysicist. I re-worked this problem last night and I might be on the verge of finding this out because I did similar to what you were saying.

This is what I did.
Assuming going down is positive
T = tension
Ĩ = Torque

Equation for Newton's law:
ΣFy = -T + mg = ma

Equation for Torque:
ΣĨ = T(r) = Iα

We know that:
Icylinder = ½mr²
α = a/r

Rewrite Torque equation:
ΣĨ = T(r) = (½mr²)(a/r)

Solve for T:
T = ½ma

Once I find T, I can plug it into Newton's equation to find the acceleration. My answer come out to be 6.53 m/s². Because there are 2 strings, I think I may have to write -2T for Newton's equation.. I'm not 100% positive.

 

[alphysicist]

Thank you for your reply alphysicist. I re-worked this problem last night and I might be on the verge of finding this out because I did similar to what you were saying.

This is what I did.
Assuming going down is positive
T = tension
Ĩ = Torque

Equation for Newton's law:
ΣFy = -T + mg = ma

Equation for Torque:
ΣĨ = T(r) = Iα

We know that:
Icylinder = ½mr²
α = a/r

Rewrite Torque equation:
ΣĨ = T(r) = (½mr²)(a/r)

Solve for T:
T = ½ma

Once I find T, I can plug it into Newton's equation to find the acceleration. My answer come out to be 6.53 m/s². Because there are 2 strings, I think I may have to write -2T for Newton's equation.. I'm not 100% positive.

That looks like the right answer to me. About the T, yes really it should be 2T in both the force equation and torque equation to match the diagram. (That is, if you had solved for T, it would not be the T in the diagram.)