Answer:
a) 1.18 seconds
b) 8.6 m
c) 5.19 revolutions
d) 6.07 m/s
Explanation:
<u>Step 1: </u>Data given
radius of the ball = 11.0 cm
Initial speed of the ball = 8.50 m/s
The coefficient of kinetic friction between the ball and the lane is 0.210.
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<em>(a) For what length of time does the ball skid?</em>
The velocity at time t can be written as v(t) = v0 + at
⇒ with v(t) = the velocity at time t
⇒ with v0 : the initial velocity = 8.50 m/s
⇒ with a = the acceleration (in m/s²)
⇒The acceleration (negative) due to friction: a = -µg
⇒ with µ = 0.210
⇒ with g = 9.81 m/s²
v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t
Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α
τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m
so α = 0.227m / 0.00484m = 46.9 rad/s²
angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t
The ball stops sliding when v(t) = ω(t) * r
8.5 - 2.06t = 46.9*0.11*t = 5.159t
7.219t = 8.5
<u>t = 1.18 seconds</u>
<em>b) How far down the lane does it skid?</em>
s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = <u>8.6 m</u>
<em>c) How many revolutions does it make before it starts to roll?</em>
The angular acceleration of the ball is:
α = τ/I
⇒ with τ = the torque experienced by the ball due the frictional force
⇒ τ = fk*R
α = fk*R /I
⇒ I = 2/5 m*R²
⇒ fk = µk*m*g
α = (µk*m*g*R)/(2/5mR²)
α = 5µk*g /2R
The angular displacement of the ball is:
∅ = 1/2αt²
⇒ The ball does not have an initial angular velocity
∅ =1/2*(5µk*g/2)*t²
∅ = 5µkgt²/4R
∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)
∅ = 32.6 rad
Number of revolutions = 32.6 rad /2π
<u>Number of revolutions = 5.19</u>
<em>(d) How fast is it moving when it starts to roll?</em>
v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = <u>6.07 m/s</u>
is the refractive index of the first medium
is the angle of incidence
is the refractive index of the second medium
is the angle of refraction
(refractive index of air), 
and 
(refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
