Rotational Kinematics of two masses

[tbdm]

Homework Statement

Consider a system of two masses joined by a massless string with the string passing over a massless frictionless pulley with a radius of
5.0 cm.
The mass of the left is
9.00 kg
and the mass on the right is
1.60 kg.
Find the angular acceleration of the pulley when the masses are released from rest, and in which direction the pulley is spinning. (Find the magnitude of the angular acceleration.)

Homework Equations

F=m*a
torque = I*alpha (angular accerleation)
T=m*g-m*a

The Attempt at a Solution

I tried getting the two tensions. T1 = (9kg)(9.8m/s^2)-(9kg)a; T2 = (1.6kg)(9.8)+(1.6kg)a
Then torque 1 - torque 2 = I*alpha, though the second part is 0 because the mass of the pulley is 0.
So I end up with Tension 1 equals tension 2 (I divided out the radius).
(9kg)(9.8m/s^2)-(9kg)a = (1.6kg)(9.8)+(1.6kg)a
This gave me a = 9.8 m/s^2. I divided it by the radius, .05 meters, to get an angular acceleration of 196 radians/second. This answer is wrong however.

 

[tiny-tim]

hi tbdm!

(9kg)(9.8m/s^2)-(9kg)a = (1.6kg)(9.8)+(1.6kg)a
This gave me a = 9.8 m/s^2.

nooo … it gives you 7.4(9.8) = 10.6a

(btw, you could have said the tensions were the same, since it's a continuous string, and the pulley has zero mass)

 

[tbdm]

Alright, it looks like I accidently switched the sign on the 1.6a when I was working it out. So that makes the tangential acceleration ≈6.842 m/s^2, and the angular acceleration is then 136.83 radians/s^2, does that sound right?

 

[hjelmgart]

Seems good to me. Check if you used the correct numbers, etc. The method is correct, at least!

 

[tiny-tim]

yup, looks good!

(except too many significant figures )

 

[tbdm]

Haha yeah, I've always found it funny that my professors tell us to use sig figs, and yet use homework programs that not only don't count them, they sometimes think the answer is wrong if you do use them. Promotes bad habits

Thank you all