Respuesta :
(a) PE is converted into KE. Measured relative to the pivot point, initial
PE = (1.20kg * 0.270m + 2.00kg * (0.0500 + 0.270)m) * 9.8m/s²
PE = 9.45 J ◄ becomes the KE at 90º
(b) KE = 9.45 J = ½Iω²
where
I = I_rod + I_sphere
I = mL²/3 + (2/5)MR² + M(R+L)² → assuming the ball is solid
and invoking the parallel axis theorem
I = 1.20kg * (0.270m)² / 3 + (2/5) * 2.00kg * (0.0500m)² + 2.00kg * (0.320m)²
I = 0.236 kg·m²
so
9.45 J = ½ * 0.236kg·m² * ω²
ω = 8.95 rad/s ◄ angular speed of rod and ball
(c) v = ωr = 8.95rad/s * (0.270 + 0.0500)m = 2.86 m/s ◄
(d) V = √(2gh) = √(2 * 9.8m/s² * 0.320m) = 2.50 m/s ◄
PE = (1.20kg * 0.270m + 2.00kg * (0.0500 + 0.270)m) * 9.8m/s²
PE = 9.45 J ◄ becomes the KE at 90º
(b) KE = 9.45 J = ½Iω²
where
I = I_rod + I_sphere
I = mL²/3 + (2/5)MR² + M(R+L)² → assuming the ball is solid
and invoking the parallel axis theorem
I = 1.20kg * (0.270m)² / 3 + (2/5) * 2.00kg * (0.0500m)² + 2.00kg * (0.320m)²
I = 0.236 kg·m²
so
9.45 J = ½ * 0.236kg·m² * ω²
ω = 8.95 rad/s ◄ angular speed of rod and ball
(c) v = ωr = 8.95rad/s * (0.270 + 0.0500)m = 2.86 m/s ◄
(d) V = √(2gh) = √(2 * 9.8m/s² * 0.320m) = 2.50 m/s ◄
The law of conservation of energy states that the total energy in a closed system is constant potential energy
(a) The rotational kinetic energy after the combination rotates through 90° is 7.868 J
(b) The angular speed of the rod and ball is approximately 8.167 rad/s
(c) The linear speed of the rod and ball is approximately 2.63 m/s
(d) The linear speed of the rod and ball combination is 0.124 m/s faster than the speed of ball free falling from 32.0 cm
Reason:
The known parameter of the system are;
Length of the cylindrical rod, l = 27.0 cm
Mass of the rod, m = 1.20 kg
Mass of ball, M = 2.00 kg
Diameter of the ball, d = 10.00 cm
(a) Required:
The rotational kinetic energy of the ball when it rotates 90°
Solution:
The total potential energy of the rod and ball at the top is the rotational kinetic energy of the ball at the bottom, [tex]K.E._{rot}[/tex]
The total potential energy of the ball and rod, [tex]P.E._{tot}[/tex] = Gravitational potential energy of the rod, [tex]P.E._{rod}[/tex] + The gravitational potential energy of the ball, [tex]P.E._{ball}[/tex]
[tex]P.E._{rod}[/tex] = 0.5 × m× g × l
∴ [tex]P.E._{rod}[/tex] = 0.5×1.20×9.81×0.27 = 1.58922
[tex]P.E._{ball}[/tex] = 2×9.81×(0.27 + 0.05) = 6.2784
[tex]P.E._{tot}[/tex] = 1.58922 J + 6.2784 J ≈ 7.868 J
- [tex]K.E._{rot}[/tex] = 7.868 J
(b) From [tex]K.E._{rot} = \dfrac{1}{2} \cdot I\cdot \omega^2[/tex], we have;
Where;
I = The moment of inertia of the combination = Moment of inertia of rod + Moment of inertia of ball
[tex]Moment \ of \ inertia \ of \ rod,\, I_{rod} = \dfrac{m\cdot L^2}{3}[/tex]
[tex]I_{rod} = \dfrac{1.20\times 0.27^2}{3} = 0.02916[/tex]
[tex]Moment \ of \ inertia \ of \ ball,\, I_{ball} =\left(\dfrac{2}{5} \right) \cdot M\cdot R^2 + M \cdot (R + L)^2[/tex]
[tex]I_{ball} =\left(\dfrac{2}{5} \right) \times 2.0\times 0.05^2 + 2.0 \times (0.050 + 0.27)^2 = 0.2068[/tex]
[tex]I = I_{rod} + I_{ball}[/tex]
I = 0.02916 kg·m² + 0.2068 kg·m² = 0.23596 kg·m²
[tex]\dfrac{1}{2} \times 0.23596 \, kg/m^2 \times \omega^2 \approx 7.868 \ kJ[/tex]
Therefore;
[tex]\omega =\sqrt{ \dfrac{7.868 \, kJ}{\dfrac{1}{2} \times 0.23596 \, kg/m^2} } \approx 8.167[/tex]
- The angular speed of the rod and ball ball, ω ≈ 8.167 rad/s
(c) The linear speed, v = ω × r
∴ v = 8.167 × (0.052 + 0.27) = 2.629774
- The linear speed of the ball = 2.629774 m/s ≈ 2.63 m/s
(d) The speed of the ball falling freely from a height of 32.0 cm is given as follows;
Speed, v = [tex]\sqrt{2 \cdot g \cdot h}[/tex]
Therefore;
[tex]v = \sqrt{2 \times 9.81 \ m/s^2 \times 0.32 \ m } \approx 2.506 \ m/s[/tex]
- The difference in speed is 2.629774 m/s - 2.506 m/s ≈ 0.123774 m/s
The linear speed following the rotation of the combination is approximately 0.124 m/s faster than the speed of the ball in free fall
Learn more here:
https://brainly.com/question/13967084
https://brainly.com/question/15296969
https://brainly.com/question/15261931
