How Is Angular Acceleration Calculated in a Rotational Torque Problem?

[hun_nomin]

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.

 

[marlon]

https://www.physicsforums.com/showthread.php?p=444214#post444214

please do not double post

marlon

 

[wais]

First, let's break down the problem into smaller parts. The first part is to find the tension in the string when the 5 kg mass is attached. We can use the equation F = ma, where F is the tension, m is the mass of the object, and a is the acceleration. Since we are looking for the tension, we rearrange the equation to T = ma. Plugging in the values, we get T = (5 kg)(9.8 m/s^2) = 49 N. This is the tension in the string when the mass is attached.

Next, we need to find the moment of inertia of the spool. The moment of inertia, represented by the symbol I, is a measure of an object's resistance to changes in its rotation. For a cylindrical spool, the moment of inertia is given by I = 1/2 * mr^2, where m is the mass of the spool and r is the radius. Plugging in the values, we get I = 1/2 * (0.5 kg)(0.075 m)^2 = 0.00140625 kg*m^2.

Now, we can use the equation for rotational torque, τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Rearranging the equation to solve for α, we get α = τ/I. Plugging in the values, we get α = (49 N)(0.075 m)/0.00140625 kg*m^2 = 2616.667 rad/s^2.

This is the angular acceleration of the spool. To find the final angular velocity, we can use the equation ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity (which is 0 in this case), and t is the time. We need to find the time it takes for the string to unwind, which can be calculated using the length of the string and the linear velocity of the 5 kg mass. The linear velocity can be calculated using v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius of the spool. Plugging in the values, we get v = (2616.667 rad/s)(0.075 m) = 196.25 m/s. The time