Question

A pitcher rotates his arm to throw a baseball. The distance from the centre of rotation of the pitcher's arm to his hand is 110 cm and remains constant over the throw. At the start of the throw, the pitcher's arm is already rotating at 0.250 rad/s and it takes him 250 ms to complete the throw. if the arc length that his hand traces over the throw is 2.70m, calculate the transverse velocity of his hand at the end of the throw.

Answer 1

GIven in question are some initials of the circular motion that was made by the pitcher. Now we don't know the value of angular acceleration, so we assume it to be .

Now the equation of motion can be written as

= t + 1/2t2

Now arc length of a circle is given as r

=> = arc/r = 2.7m/1.1m = 27/11

Now using the value of and time interval t = 250 ms = 0.250s, = 0.250 rad/s and  in above equation we get,

27/11 = 0.250 x 0.250 + 1/2 (0.250)2

=> = 842/11 = 76.54 rad/s

Now we move to find final angular velocity,

= + t

=> = 0.250 + 76.54 x 0.250s

=> = 19.386 rad/s

So the final transverse velocity will be equal to vT = x r = 19.386 rad/s x 1.1m = 21.325 m/s