The rotational Kinetic energy for the ball is 5.344 J.
The ball's total kinetic energy just before it hits the floor is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy is given by
KE₁ = (1/2)mv²
where m is the mass of the ball and v is its velocity. In this case, m = 0.2 kg and v = 5 m/s. Therefore, the translational kinetic energy is:
KE₁ = (1/2)(0.2 kg)(5 m/s)² = 2.5 J
The rotational kinetic energy is given by:
KE₂ = (1/2)Iω²
where I is the moment of inertia of the ball and ω is its angular velocity. The moment of inertia of a solid ball is (2/5)mr², where r is the radius of the ball. In this case, r = 0.1 m. Therefore, the rotational kinetic energy is:
KE₂ = (1/2)((2/5)(0.2 kg)(0.1 m)²)ω² = (1/50)ω² J
We can relate the angular velocity ω to the linear velocity v using the equation:
v = rω
In this case, v = 5 m/s and r = 0.1 m. Therefore, ω = 50 rad/s. Substituting this into the equation for rotational kinetic energy, we get:
KE₁= (1/50)(50 rad/s)² = 5 J
Therefore, the total kinetic energy of the ball just before it hits the floor is:
KE = KE₂ + KE₁ = 2.5 J + 5 J = 7.5 J
The gravitational potential energy of the ball just before it hits the floor is given by:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the floor. In this case, m = 0.2 kg, g = 9.8 m/s², and h = 1.1 m. Therefore, the gravitational potential energy is:
PE = (0.2 kg)(9.8 m/s²)(1.1 m) = 2.156 J
The law of conservation of energy states that the total energy of an isolated system remains constant. Therefore, the total energy of the ball just before it hits the floor is equal to the total energy of the ball when it is released from the table. The total energy of the ball when it is released from the table is equal to its kinetic energy, which we have already determined to be 7.5 J. Therefore, the rotational kinetic energy of the ball just before it hits the floor is:
KE_rotational = KE - PE= 7.5 J - 2.156 J = 5.344 J