Respuesta :
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×[tex]10^{-4}[/tex] m
dynamic viscosity = 1.75 ×[tex]10^{-5}[/tex] Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ [tex]\frac{du}{dy}[/tex]
so
= µ [tex]\frac{v}{h}[/tex] ............1
put here value
= 1.75×[tex]10^{-5}[/tex] × [tex]\frac{v}{10^{-4}}[/tex]
= 0.175 v
and
area between air and puck is given by
Area = [tex]\frac{\pi }{4} d^{2}[/tex]
area = [tex]\frac{\pi }{4} 0.1^{2}[/tex]
area = 7.85 × [tex]\frac{v}{10^{-3}}[/tex] m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × [tex]10^{-3}[/tex]
force = 1.374 × [tex]10^{-3}[/tex] v
and now apply newton second law
force = mass × acceleration
- force = [tex]mass \frac{dv}{dt}[/tex]
- 1.374 × [tex]10^{-3}[/tex] v = [tex]0.03 \frac{0.9v - v }{t}[/tex]
t = [tex] \frac{0.1 v * 0.03}{1.37*10^{-3} v}[/tex]
time = 2.18
so time required after impact for a puck is 2.18 seconds
