A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un
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Answer:
1) The magnitude of the angular acceleration = 67.92 rad/
2) Magnitude of the linear acceleration = 2.744 m/
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 = 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
= μ N , N = m g
= 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
torque τ = I α
τ = F R
I α = F R
(2/5) m 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/
2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane
F = - ,
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/
3) The bowling ball takes time to begin rolling without slipping:
The linear speed, =
- a t
=
- μ g t
the angular speed, ω = ω0 + α t
ω = ω0 + (5 μ g/2R ) t
= ω R
- μ g t = ω0 R + (5 μ g/2R ) t R
7 μ g t/2 = + ω0 R
hence,
t = (2 + ω0 R)/ 7 μ g
ω0 = 0 (no initial spin), therefore
t = 2 / 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 = t - (1/2) a
= (8.7) (0.906) - (1/2) (2.744)
= 6.75 m
5) The final velocity
=
- a t
= 8.7 - (2.744) (0.906)
= 6.21 m/s