How Do You Calculate Angular Acceleration and Displacement of a Rolling Ball?

[BlakeEdwards]

A solid ball with a diameter of 22 cm rolls along a horizontal surface as shown without slipping at a constant acceleration. It initially (t=0) has an angular velocity of 11 rad/s and 10 seconds later has an angular velocity of 3 rad/s.
(a)What is its angular acceleration? rad/s2
(b)What is the linear (translational) displacement of the ball? m
(c)How long (from t=0) will it take to stop? sec


I got -.8 rad/s^2
15.4 m
and 13.75 seconds

Im so confused, can someone help me please?

 

[tiny-tim]

Welcome to PF!

Hi BlakeEdwards! Welcome to PF!

(try using the X² tag just above the Reply box )

erm … you've used diameter instead of radius.

 

[kerimek]

To calculate the angular acceleration of the ball, we can use the formula: angular acceleration (α) = change in angular velocity (ω) / change in time (t). In this case, the change in angular velocity is 11 rad/s - 3 rad/s = 8 rad/s. The change in time is 10 seconds. Therefore, the angular acceleration is 8 rad/s / 10 s = 0.8 rad/s^2.

To find the linear displacement of the ball, we can use the formula: linear displacement (s) = angular displacement (θ) x radius (r). The angular displacement can be calculated by using the formula: angular displacement (θ) = initial angular velocity (ω0) x time (t) + 1/2 x angular acceleration (α) x time^2 (t^2). In this case, the initial angular velocity is 11 rad/s, the time is 10 seconds, and the angular acceleration is 0.8 rad/s^2. Therefore, θ = 11 rad/s x 10 s + 1/2 x 0.8 rad/s^2 x (10 s)^2 = 110 rad. The radius of the ball is half of its diameter, so it is 11 cm or 0.11 m. Therefore, the linear displacement is 110 rad x 0.11 m = 12.1 m.

To calculate the time it takes for the ball to stop, we can use the formula: final angular velocity (ω) = initial angular velocity (ω0) + angular acceleration (α) x time (t). In this case, the final angular velocity is 0 rad/s, the initial angular velocity is 11 rad/s, and the angular acceleration is -0.8 rad/s^2 (since the ball is decelerating). Therefore, 0 rad/s = 11 rad/s + (-0.8 rad/s^2) x t. Solving for t, we get t = 11 rad/s / 0.8 rad/s^2 = 13.75 seconds.

Therefore, the angular acceleration is 0.8 rad/s^2, the linear displacement is 12.1 m, and it will take 13.75 seconds for the ball to stop.