Question

A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg. (a) What is the angular velocity of the arm immediately after catching the ball as measured from the arm socket? (b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?

Answer 1

Answer

given,

speed of baseball = 40 m/s

mass of the baseball = 0.145 Kg

catcher arm length = 0.5 m

mass = 4 Kg

using conservation of momentum

$L_i = L_f$

$m r v =(I_1+I_2)\omega_f$

$\omega_f=\dfrac{mrv}{I_1+I_2}$

$\omega_f=\dfrac{0.5\times 0.145\times 40}{\dfrac{1}{3}4 r^2+\dfrac{1}{3}4 r^2}$

$\omega_f=\dfrac{0.5\times 0.145\times 40}{\dfrac{1}{3}\times 4\times 0.5^2+\dfrac{1}{3}\times 4 \times 0.5^2}$

$\omega_f= 6.277\ rad/s$

2) torque

$\tau = \dfrac{dL}{dt}$

L = I ω =$6.277 \dfrac{1}{3}\times 4 \times 0.5^2}$ = 2.09 kg.m^2/s

$\tau = \dfrac{0-2.09}{0.3}$

 τ = 6.97 N.m