Aspherical 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?

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!

he kicked it 10 feet straight up into the air? i think this works.

pe = weightxheight = 2000x1000=2,000,000joules