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A spherical bowling ball with mass m = 3.4 kg and radius R = 0.113 m is thrown down the lane with an initial speed of v = 8.1 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.31. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.' 1.What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

2.What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

3.How long does it take the bowling ball to begin rolling without slipping?
4.How far does the bowling ball slide before it begins to roll without slipping?
5.What is the magnitude of the final velocity?

Respuesta :

whitneytr12

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!

onyebuchinnaji

(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.

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 0.31 x 9.8)/(2 x 0.113)

α = 67.21 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 0.113 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]

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 = 8/8.18

t = 0.98 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 x 0.98) - (0.5 x 8.18 x 0.98²)

xf = 3.91 m

Final velocity of the bowling ball

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

Learn more about angular acceleration here: https://brainly.com/question/25129606

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