Answer:
1. α = 67.28 rad/s²
2. a = -3.04 m/s²
3. t = 0.76 s
4. x = 5.28 m
5. vf = 5.78 m/s
Explanation:
1. Let's use the torque definition: τ = Iα.
The inertial moment of a sphere is I = (2/5)*m*R²
And we know that the torque is the cross product between force and distance, so we would have τ = FxR=|F|*|R|*sin(90)=|F|*|R|=μ*mg*R
Using these two definitions, we have: (2/5)*m*R²*α = μ*mg*R
So the magnitude of the angular acceleration would be: α = (5/2R)*μ*g = 67.28 rad/s².
2. The force definition is F = m*a, when a is the linear acceleration.
F = -μ*mg.
Then -μ*mg = m*a. Solving the equation for a we have: a = -μ*g = -3.04 m/s².
3. To get the time when the ball star to rolling we need to use angular and linear velocity equation.
- ωf = ω0 + α*t ; we assume that initial angular velocity is 0.
- vf = v0 - a*t; v0 is the initial linear velocity
The relation to pure rolling is: v = ω*R. Rewriting this equation in terms of time v0 - a*t = α*t*R, so t = v0/(α*R+a) = 0.76 s.
4. Using the distance equation: xf = x0 + v0*t - 0.5*a*t² = 5.28 m.
5. vf = v0 - a*t = 5.78 m/s.
Have a nice day!
(a) The angular acceleration of the bowling ball is 67.21 rad/s².
(b) The linear acceleration of the bowling ball is 8.18 m/s².
(c) The time taken for the bowling ball to begin tolling without slipping is 0.98 s.
(d) The distance traveled by the bowling ball is 3.91 m.
(e) The final velocity of the bowling ball is 0.18 m/s.
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 0.31 x 9.8)/(2 x 0.113)
α = 67.21 rad/s²
The linear acceleration of the bowling ball is calculated as follows;
a = αR
a = 67.21 rad/s² x 0.113 m
a = 7.6 m/s²
ac = -μg
ac = -0.31 x 9.8 = -3.038 m/s²
[tex]a_t = \sqrt{a^2 + a_c^2} \\\\a_t = \sqrt{7.6^2 + (-3.038)^2} \\\\a_t = 8.18 \ m/s^2[/tex]
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 = 8/8.18
t = 0.98 s
The distance traveled by the bowling ball is calculated as follows;
xf = ut - ¹/₂at²
xf = (8 x 0.98) - (0.5 x 8.18 x 0.98²)
xf = 3.91 m
The final velocity of the bowling ball is calculated as follows;
vf² = v² + 2(-a)s
vf² = 8² - 2(8.18)(3.91)
vf²= 0.032
vf = √0.032
vf = 0.18 m/s
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