In the figure below, a wheel of radius 0.60 m is mounted on a frictionless horiz
ID: 1986583 • Letter: I
Question
In the figure below, a wheel of radius 0.60 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.40 kg · m2. A massless cord wrapped around the wheel's circumference is attached to a 5.0 kg box. The system is released from rest.Explanation / Answer
very similar.helps? --> a wheel of radius 0.467 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.296 kg·m^2. A massless cord wrapped around the wheel's circumference is attached to a 4.00 kg box. The system is released from rest.When the box has a kinetic energy of 8.00 J, what are (a) the wheel's rotational kinetic energy and (b) the distance the box has fallen? Assume free-fall acceleration to be equal to 9.81 m/s^2. soln.The speed of the block = linear speed on the rim of the wheel so 1/2*m*v^2 = 8.00j so v = sqrt(2*8.0/4.00) = 2.00m/s Therefore K wheel = 1/2*I*?^2 = 1/2*I*(v/r)^2 = 1/2*0.296*(2/0.467)^2 = 2.71J b) Now conservation of energy states (K + U) 1 = (K + U)2 but K1 = 0 and U1 = 0 so K2 + U2 = 0 8.00 + 2.71 + m*g*(-y) = 0 or y = 10.71/(4.00*9.81)= 0.273m