Respuesta :
Answer:
1) The magnitude of the angular acceleration = 67.92 rad/[tex]s^{2}[/tex]
2) Magnitude of the linear acceleration = 2.744 m/[tex]s^{2}[/tex]
3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s
4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m
5) the final velocity is 6.21 m/s
Explanation:
the given information :
Bowling mass m = 3.6 kg
Radius = 0.101 m
Initial speed [tex]v_{0}[/tex] = 8.7 m/s
Coefficient of kinetic friction μ = 0.28
1) he magnitude of the angular acceleration of the bowling ball is
F = m a
[tex]F_{g}[/tex] = μ N , N = m g
[tex]F_{g}[/tex] = μ m g
1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:
momen inersia of Bowling ball I = (2/5) m [tex]R^{2}[/tex]
torque τ = I α
τ = F R
I α = F R
(2/5) m [tex]R^{2}[/tex] α = μ m g R
α = (5 μ g / 2R) μ g R
= (5 x 0.28 x 9.8/ 2 x 0.101)
= 67.92 rad/[tex]s^{2}[/tex]
2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane
F = - [tex]F_{g}[/tex] , [tex]F_{k}[/tex] is the force of kinetic friction
m a = - μ m g, remove m
the magnitude of linear accelaration is
a = μ g
= (0.28) (9.8)
= 2.744 m/[tex]s^{2}[/tex]
3) The bowling ball takes time to begin rolling without slipping:
The linear speed, [tex]v_{t}[/tex] = [tex]v_{0}[/tex] - a t
[tex]v_{t}[/tex] = [tex]v_{0}[/tex] - μ g t
the angular speed, ω = ω0 + α t
ω = ω0 + (5 μ g/2R ) t
[tex]v_{t}[/tex] = ω R
[tex]v_{0}[/tex] - μ g t = ω0 R + (5 μ g/2R ) t R
7 μ g t/2 = [tex]v_{0}[/tex] + ω0 R
hence,
t = (2 [tex]v_{0}[/tex] + ω0 R)/ 7 μ g
ω0 = 0 (no initial spin), therefore
t = 2 [tex]v_{0}[/tex] / 7 μ g
= 2 x 8.7 / 7 (0.28) (9.8)
= 0.906 s
4) How long it takes for the bowling ball to begin rolling without slipping, S
S = [tex]v_{0}[/tex] t - (1/2) a [tex]t^{2}[/tex]
= (8.7) (0.906) - (1/2) (2.744) [tex]0.906^{2}[/tex]
= 6.75 m
5) The final velocity
[tex]v_{t}[/tex] = [tex]v_{0}[/tex] - a t
[tex]v_{t}[/tex] = 8.7 - (2.744) (0.906)
[tex]v_{t}[/tex] = 6.21 m/s
(a) The angular acceleration of the bowling ball is 67.92 rad/s².
(b) The linear acceleration of the bowling ball is 7.39 m/s².
(c) The time taken for the bowling ball to begin tolling without slipping is 1.18 s.
(d) The distance traveled by the bowling ball is 5.12 m.
(e) The final velocity of the bowling ball is 0.13 m/s.
Conservation of angular momentum
The angular acceleration of the bowling ball can be determined by applying the principle of conservation of angular momentum as follows;
Iα = FR
²/₅mR²α = (μmg)R
²/₅Rα = μg
2Rα = 5μg
α = 5μg/2R
α = (5 x 0.28 x 9.8)/(2 x 0.101)
α = 67.92 rad/s²
Linear acceleration of the bowling ball
The linear acceleration of the bowling ball is calculated as follows;
a = αR
a = 67.21 rad/s² x 0.101 m
a = 6.86 m/s²
ac = -μg
ac = -0.28 x 9.8 = -2.7 m/s²
[tex]a_t = \sqrt{a ^2 + a_c^2} \\\\a_t = \sqrt{6.86^2 + (-2.74)^2} \\\\a_t = 7.39 \ m/s^2[/tex]
Time taken for the bowling ball to begin rolling
The time taken for the bowling ball to begin tolling without slipping is calculated as follows;
v = u - at
0 = u - at
at = u
t = u/a
t = 8.7/7.39
t = 1.18 s
Distance traveled by the bowling ball before slipping
The distance traveled by the bowling ball is calculated as follows;
xf = ut - ¹/₂at²
xf = (8.7 x 1.18) - (0.5 x 7.39 x 1.18²)
xf = 5.12 m
Final velocity of the bowling ball
The final velocity of the bowling ball is calculated as follows;
vf² = v² + 2(-a)s
vf² = 8.7² - 2(7.39)(5.12)
vf²= 0.0164
vf = √0.0164
vf = 0.13 m/s
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