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------------------------------------------------ A solid ball with a mass of 2kg and the radius of 8 cm rolls without slipping down a ramp of total height 1.5m. What would be its translational velocity at the bottom?

Using conservation of energy, the gravitational potential energy at the top is converted into both translational and rotational kinetic energy at the bottom.

By solving the energy equations, we find that the translational velocity of the ball at the bottom of the ramp is approximately 5.32 m/s.

We need to use the conservation of energy principle. At the top of the ramp, the ball has potential energy, and at the bottom, this energy is converted into both translational and rotational kinetic energy.

Step-by-step explanation:

Calculate the gravitational potential energy (GPE) at the top:
GPE = mgh
where m = 2 kg, g = 9.8 m/s², h = 1.5 m.
GPE = 2 kg × 9.8 m/s² × 1.5 m = 29.4 J
At the bottom, the potential energy is converted into translational kinetic energy (TKE) and rotational kinetic energy (RKE).
The translational kinetic energy (TKE) is given by:
TKE = 1/2 mv²
The rotational kinetic energy (RKE) for a solid sphere is:
RKE = 1/2 Iω²
The moment of inertia (I) for a solid sphere is:
I = 2/5 mR²
where R = 0.08 m (converting 8 cm to meters).
I = 2/5 × 2 kg × (0.08 m)² = 0.00512 kg·m²
Since the ball rolls without slipping, ω = v/R, where v is the translational velocity and R is the radius.
RKE = 1/2 × 0.00512 kg·m² × (v/0.08 m)² = 0.04v² J
Using the conservation of energy, we have:
GPE = TKE + RKE
29.4 J = 1/2 × 2 kg × v² + 0.04v²
29.4 J = v² + 0.04v²
29.4 J = 1.04v²
v² = 29.4 J / 1.04
v² = 28.27
v ≈ 5.32 m/s