How Do You Calculate Angular Acceleration in a Rotational Torque Problem?

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To calculate the angular acceleration of a cylindrical spool with a mass attached, first apply Newton's second law to the hanging mass, considering the forces acting on it. The tension in the string must be determined to find the angular acceleration, which is related to the tangential acceleration of the spool. The moment of inertia (I) for the spool is calculated as 1/2 * mass * radius^2. Once the tension (T) is found, it can be substituted back into the equations to solve for angular acceleration and the final rotational speed of the spool. The discussion emphasizes the importance of correctly calculating tension before proceeding with angular acceleration.
hun_nomin
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Question:

A light string 4 m long is wrapped around a cylindrical spool with a radius of 0.075 m and a mass of .5 kg. A 5 kg mass is then attached to the free end of the string causing the string to unwind from the spool.

a) What is the angular accelerataion of the spool?

b) how fast will the spool be rotating after all of the string has unwound?

I got an answer of 2613.333 for (a), but my teacher claims the answer is wrong. I just want to solve this problem or else it's going to kill me...so if any of you know how, please reply. He said the tension must be solved first. but i have no idea. i though the time would be .9 but seems not.
 
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hun_nomin said:
Question:

A light string 4 m long is wrapped around a cylindrical spool with a radius of 0.075 m and a mass of .5 kg. A 5 kg mass is then attached to the free end of the string causing the string to unwind from the spool.

a) What is the angular accelerataion of the spool?

b) how fast will the spool be rotating after all of the string has unwound?

I got an answer of 2613.333 for (a), but my teacher claims the answer is wrong. I just want to solve this problem or else it's going to kill me...so if any of you know how, please reply. He said the tension must be solved first. but i have no idea. i though the time would be .9 but seems not.

Let's say that the string hangs on the right side of the spool, ok ?

Suppose the y-axis is upward vertically. First we apply Newton's second law for the mass. The acceleration is negative since it moves downward:

5*a = 5*g - T and T is the tension in the string.

Then we apply the second law of Newton for the spool with rotational inertia denoted as I (you can look this up) and angular acceleration a' : Ia' = torque on spool

the torque coming from the string-force (tangential to the spool) is equal to -RT (- because the spool moves along with the clock and R is the radius of the spool).

now you know that tangential acceleration is connected to the angular acceleration via a = a' * R (a' < 0 because the spool moves clockwise and a also < 0)

Now we have that Ia' = -RT or Ia/R = -RT or a = -R²T/I

let's put this in the first equation for the mass

5 (-R²T/I) = 5 * g - T

calculate T from this equation

once you know T, you know a...

marlon
 
Last edited:
Thanks

Thanks...i'm going to go solve it now. And sorry for double topics...i didn't realize i was posting a high school problem in the college section so i decided to post it again. Sorry
 
What is I?

I'm not sure what I is... is it equal to 1/2(.5).075^2?
 
hun_nomin said:
I'm not sure what I is... is it equal to 1/2(.5).075^2?

that is correct !

marlon
 
T

If i did it correctly i got a T of 1.96. Is that the right answer?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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