Question

Integrated Concepts When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity? (b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s? (c) Find the maximum range of the football, neglecting air resistance.

Answer 1

Answer:

(a) 33.3 rad/s

(b) 500 N

(c) 40.8 m

Explanation:

Angular velocity $\omega$  is given by

$\omega=\frac {v}{r}$

Where v is linear velocity and r is radius of circle

$\omega=\frac {35 m/s}{1.05m}=33.3 rad/s$

Angular velocity is 33.3 rad/s

(b)

Velocity, v is calculated as

$V=\frac {r}{t}$  where r is radius of circle and t is time

Making r the subject of the formula

r=vt

In this case, t=20 ms converted to seconds will be 20/1000=0.02 s

r=20*0.02=0.4m

Force, $F=ma_{c}$  where m is mass and $a_{c}$  is centripetal acceleration

$a_{c}=\frac {v^{2}}{r}$  

Therefore

$F=m\frac {v^{2}}{r}$  

$F=05\frac {20^{2}}{0.4}=500N$

Force=500 N

(c)

Maximum range covered by the football

$d=\frac {v^{2}}{g}sin(2\theta)$  where g is gravitational constant taken as 9.8, $\theta$  is 45 and v is 20 m/s

$d=\frac {20^{2}}{9.8}sin((2)(45))=40.8 m$

Maximum range covered is 40.8 m