In the figure below, a wheel of radius 0.10 m is mounted on a frictionless horiz
ID: 1984098 • Letter: I
Question
In the figure below, a wheel of radius 0.10 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.
(a) When the box has a kinetic energy of 6.0 J, what is the wheel's rotational kinetic energy?
(b) What is the distance the box has fallen?
In the figure below, a wheel of radius 0.10 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. (a) When the box has a kinetic energy of 6.0 J, what is the wheel's rotational kinetic energy? (b) What is the distance the box has fallen?Explanation / Answer
a) KE of box = 6
Hence, 0.5m*v2 = 6
v = 1.55 m/s
Hence, the circumferential velocity of pulley = 1.55 m/s
So, angular velocity of pulley = 1.55/r
hence, Rotational KE of pulley = 0.5*I*(1.55/r)2
Calculate it for r = 0.1 m
b) Total KE of the system is due to the loss in potential energy of the box.Hence,
0.5*I*(1.55/r)2 + 6 = 5*g*H
where, H is the height fallen
Do the calculations