Question

a 0.56kg yo-yo is released from rest and allowed to drop. the moment of inertia of the yo-yo is 2.9x10^-5 kg m^2 and the radius of the axle is 0.0064m. what is the linear speed of the yo-yo after it has dropped through a height of 0.5

Answer 1

2.5 m/s

Step-by-step explanation:

Energy is conserved.

Potential energy = Kinetic energy + Rotational energy

mgh = ½ mv² + ½ Iω²

mgh = ½ mv² + ½ I (v/r)²

mgh = ½ v² (m + I/r²)

(0.56 kg) (9.8 m/s²) (0.5 m) = ½ v² [0.56 kg + ½ (2.9×10⁻⁵ kg m²) / (0.0064 m)²]

2.744 Nm = ½ v² (0.56 kg + 0.354 kg)

2.744 Nm = ½ v² (0.914 kg)

v² = 6.00 m²/s²

v = 2.5 m/s