Question

A wheel with a weight of 387 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 24.7 rad/s . The radius of the wheel is 0.592 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3454 J .

Calculate h.

Use 9.81 m/s2 for the acceleration due to gravity.

Answer 1

Initial kinetic energy = rotational kinetic energy + translational kinetic energy

FOR rolling without slipping, V = RW WHERE R IS RADIUS AND W IS ANGULAR VELOCITY

Initial kinetic energy = 1/2*0.8MR² *W² + 1/2*M*(RW)²

Initial kinetic energy = 0.4MR² *W² + 0.5M*R²W²

Initial kinetic energy = 0.9M*R²W²

Now using conservation of energy

0.9M*R²W² + ( - 3454) = Mgh

Notice the negative sign because work done by friction is negative

Here M = W /g = 387 / 9.8 = 39.489 kg

0.9*39.489*0.592²*24.7² - 3454 = 387*h

h = 0.9*39.489*0.592²*24.7² - 3454 / 387

h = 4144.99 / 387

h = 10.6909 m