Angular acceleration of the disk as a function of time

[iamwilson]

Homework Statement

A disk of radius 21.0cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (the figure ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation a(t)=At, where t is in seconds and A is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.60m/s^2 .

A) find A
B)Express the angular acceleration of the disk as a function of time.
c)How much time after the disk has begun to turn does it reach an angular speed of 12.0rad/s
Through what angle has the disk turned just as it reaches 12.0rad/s ? (Hint: See Section 2.6 in the textbook.)

Homework Equations

i solved a, b, &c, I'm stuck on part d

The Attempt at a Solution

for A, i got 0.533m/s^2, b)2.54rad/s^3(t), c)3.07s, d)?
for part d, I'm clueless on which formula to use

 

[iamwilson]

NVM, i figured it out

 

[Moetasim]

, can you please provide guidance on how to approach this problem?

For part d, you can use the equation for angular velocity:
ω(t) = ω0 + αt
where ω is the angular velocity at time t, ω0 is the initial angular velocity (in this case, 0), and α is the angular acceleration.

Since you know that the angular acceleration is 2.54rad/s^3(t), you can substitute that into the equation:
ω(t) = 2.54rad/s^3(t)t

To find the time when the angular velocity reaches 12.0rad/s, you can set ω(t) = 12.0rad/s and solve for t:
12.0rad/s = 2.54rad/s^3(t)t
t = 12.0rad/s / 2.54rad/s^3
t = 4.72s

So, it takes 4.72 seconds for the disk to reach an angular velocity of 12.0rad/s.

To find the angle that the disk has turned at this point, you can use the equation for angular displacement:
θ(t) = θ0 + ω0t + 1/2αt^2
where θ is the angular displacement at time t, θ0 is the initial angular displacement (in this case, 0), ω0 is the initial angular velocity (in this case, 0), and α is the angular acceleration.

Substituting in the values that you know, you get:
θ(t) = 0 + 0 + 1/2(2.54rad/s^3)(4.72s)^2
θ(t) = 28.4rad

So, the disk has turned 28.4 radians just as it reaches an angular velocity of 12.0rad/s.

 

[Moetasim]

.

I would like to commend you on your progress in solving parts A-C of the problem. It shows that you have a strong understanding of the concepts and equations involved. For part D, you can use the equation ω = ω0 + αt, where ω is the final angular velocity (12.0rad/s), ω0 is the initial angular velocity (0), α is the angular acceleration (2.54rad/s^3), and t is the time it takes to reach the final angular velocity (unknown). You can rearrange the equation to solve for t, and then plug in the values to find the time it takes for the disk to reach an angular speed of 12.0rad/s.

To find the angle the disk has turned, you can use the equation θ = ω0t + 1/2αt^2, where θ is the angle turned, ω0 is the initial angular velocity (0), α is the angular acceleration (2.54rad/s^3), and t is the time it takes to reach the final angular velocity (3.07s). Again, you can rearrange the equation to solve for θ and then plug in the values to find the angle the disk has turned just as it reaches 12.0rad/s.

Keep up the good work and don't be afraid to ask for help or clarification when needed. Science is all about collaboration and seeking knowledge together.