# Angular acceleration problem

1. Mar 29, 2005

### Punchlinegirl

M, a solid cylinder (M=1.99 kg, R=0.133 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force of F which equals the weight of a 0.830 kg mass, i.e, F=8.142 N. Calculate the angular acceleration of the cylinder.

I tried drawing a free body diagram and setting the forces equal to ma.
F_n- mg-F_t= Ma_y
since a= 0, F_n= Mg + F_t
torque= RF_t sin 90= Ia
RF_t=Ia
I=(1/2)MR^2
F_t= (1/2)MRa
mg-ma=(1/2)MR(a/R)
(1/2M +m)a= mg
a=mg/ (1/2 M=m)
I plugged in my numbers, but I think the equation is wrong. Help?

2. Mar 29, 2005

### xanthym

From the problem statement:
{String Tension} = S
{Mass of Suspended Entity} = m = (0.830 kg)
{Weight of Mass} = W = (0.830 kg)*(9.81 m/sec^2) = (8.1423 N)
{Cylinder Mass} = M = (1.99 kg)
{Cylinder Radius} = R = (0.133 m)
{Cylinder Moment of Inertia} = I = (1/2)*M*R^2
{Cylinder Angular Acceleration} = α

For the suspended entity:
{Net Force} = ma =
= W - S
::: ⇒ S = W - ma ::: Eq #1

For the cylinder:
{Net Torque} = Iα = I*a/R =
= S*R
::: ⇒ S = I*a/R^2 ::: Eq #2

Equating Eq #1 and Eq #2:
W - ma = I*a/R^2
::: ⇒ a = W/{m + I/R^2}
::: ⇒ a = W/{m + (1/2)*M*R^2/R^2}
::: ⇒ a = W/{m + (1/2)*M}
::: ⇒ a = (8.1423 N)/{(0.830 kg) + (1/2)*(1.99 kg)}
::: ⇒ a = (4.46153 m/sec^2)
::: ⇒ α = a/R = (4.46153 m/sec^2)/(0.133 m) = (33.545 radians/sec^2)

~~

3. Mar 29, 2005

### Staff: Mentor

Your first error is in assuming that the string exerts a force equal to the weight of the hanging mass. (If the force were equal, then the mass would be in equilibrium.) But it looks like you don't use this fact later on.

For the cylinder, the only forces you need consider are those that create a torque. Thus the normal force (supporting the cylinder) and the weight of the cylinder are irrelevant; all that counts is the force F that the string exerts.

Applying Newton's 2nd law:
To the cylinder:
$\tau = I\alpha$ --> $F R = I\alpha$​
To the hanging mass:
$mg - F = ma$​

So far, so good. (a is really alpha)
All good. I assume you meant to write: $a = mg/(M/2 + m)$.
Did you remember to convert to angular acceleration?

4. Mar 30, 2005

### Punchlinegirl

ok, i used the equation and tried converting to angular acceleration by using
a=alpha* r
and got the angular acceleration to be 3.87 rads/s^2, which isn't right...

5. Mar 30, 2005

### whozum

Alpha is the angular acceleration, you are using your equation backwards :). Solve it for alpha and try it that way.

6. Apr 7, 2005

### trisha320

i have a similar problem to punchlinegirl, and i was wondering how do i find alpha from angular acceleration

7. Apr 7, 2005

### xanthym

Angular acceleration IS "α". You might be thinking of the following relationships:
{Angular Acceleration} = α =
= {Linear Acceleration}/R =
= a/R
{Angular Velocity} = ω =
= {Linear Velocity}/R =
= v/R

Also:
{Angular Velocity} = ω =
= 2*π*{Rotation Frequency} =
= 2*π*f

~~

8. Apr 7, 2005

### kishin7

M=1.99 kg, R=0.133 m, F=8.142 N

I = .5MR^2, T(1) = r x F and T(2) = I x alpha.

Solve for I and then solve T(1) using F given and then once you have both plug into T(2).

Last edited: Apr 7, 2005
9. Apr 10, 2005

### Felix83

The force is equal to the weight of a hanging mass. This doesnt mean that there is actually a mass hanging off of it, it means there is a constant force applied of 8.142N.
Once you see that the problem is simple.
You have the equation T=I*alpha
alpha is angular acceleration which is what you need so just solve for alpha.
alpha=T/I where T is the torque applied and I is the moment of inertia.
T = F*r (when torque is perpendicular)
T = (8.142N) (0.133m)
T= 1.083Nm

Now find I, since it is a solid cylinder
I=(1/2)Mr^2
I = (1/2) (1.99kg) (0.133m)^2
I = 0.0176 kg*m^2

now recall
alpha = T/I
alpha = (1.083Nm)/(.0176kg*m^2)