Respuesta :
Answer:
a) 1.18 seconds
b) 8.6 m
c) 5.19 revolutions
d) 6.07 m/s
Explanation:
Step 1: 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.
(a) For what length of time does the ball skid?
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
t = 1.18 seconds
b) How far down the lane does it skid?
s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m
c) How many revolutions does it make before it starts to roll?
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
Number of revolutions = 5.19
(d) How fast is it moving when it starts to roll?
v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s
