Respuesta :
Answer:124.95 rad/s
Explanation:
Given
radius of ball [tex] r=22 cm[/tex]
coefficient of friction [tex]\mu =0.2[/tex]
ball center has a velocity of [tex]v=11 m/s[/tex]
back spin angular velocity is [tex]\omega _0[/tex]
mass of ball [tex]m=5 kg[/tex]
as there is no motion in Y direction therefore net Force is zero
[tex]\sum F_y=0[/tex]
[tex]W-N=0[/tex]
[tex]W=N[/tex]
Friction will provide the torque
[tex]f_r=\mu N[/tex]
motion is in x direction therefore
[tex]f_r=ma[/tex]
[tex]\mu N=\mu N=ma[/tex]
[tex]a=\frac{0.2\times 5\times 9.8}{5}[/tex]
[tex]a=1.96 m/s^2[/tex]
time taken to stop its linear velocity
[tex]v=u+at [/tex]
[tex]0=11-1.96\times t[/tex]
[tex]t=\frac{11}{1.96}=5.61 s[/tex]
torque [tex]f_r\cdot r=I\alpha [/tex]
[tex]I=\frac{2}{5}mr^2[/tex]
[tex]f_r\cdot r=\frac{2}{5}mr^2\cdot \alpha [/tex]
[tex]\alpha =22.27 rad/s^2[/tex]
using
[tex]\omega =\omega _0+\alpha t[/tex]
[tex]0=\omega -22.27\times 5.61[/tex]
[tex]\omega _0=124.95 rad/s[/tex]
