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!
