angular acceleration?

Question

During a tennis serve, a racket is given an angular acceleration of magnitude 125 rad/s2. At the top of the serve, the racket has an angular speed of 13 rad/s. If the distance between the top of the racket and the shoulder is 1.0 m, find the magnitude of the total acceleration of the top of the racket.

Answer 1

Let A be angular acceleration then
A=a/R
where a - tangential acceleration
R - the radius of motion in question
then
a=AR=125 x 1.0=125m/s^2

Answer 2

The formulas are quite similar: linear acceleration = (change in linear velocity) / (time period) angular acceleration = (change in angular speed) / (time period) tangential acceleration = (change in tangential speed) / (time period) OK, so now you may be asking, what is meant by "angular speed" and "tangential speed" ? Angular speed may be specified in revolutions per minute or radians per second, etc. Some angular measure per unit time. You may not remember 45 RPM plastic records that played music by rotating on a turntable, but as the "45 RPM" suggests, the turntable would rotate at 45 revolutions per minute. Now let's say one of these discs had a 3-inch radius. The "tangential speed" at the edge of the disc would be found from knowing that the circumference is 2 pi r = 6 pi inches, about 19 inches. So the tangential speed at the edge of one of these discs would be 19 inches/rev times 45 rev/min = 855 inches per minute.

Answer 3

β = dω/dt = 125 /s²
ω = 13 /s
R = 1m

Answer:
a = √[(βR)² + (ω²R)²] = R√[β² + (ω²)²]= 210 m/s²