A wheel with a weight of 394 N comes off a moving truck and

A wheel with a weight of 394 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 23.8 rad/s . The radius of the wheel is 0.589 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 3502 J .

Solution

Although it has not been specified, I assume the height h above the bottom of the hill is to be determined.

We will use the principle of conservation of energy to form an equation involving h and then resolve to determine its value. We will choose the bottom of the hill as the reference level and it needs to be understood that the wheel will have only KE at the bottom which will get exhausted by friction in part and the remaining will turn into PE.

Also, for a body rotating with Inertia I, the KE is given as 0.5 [Iw^2 + mv^2] where w is angular velocity while v is the linear one.

We can rewrite the above as: KE = 0.5 [0.800 MR^2 x 566.44  + M x 196.51]

Now, this will be equal to the sum of final PE and the work done by friction

Hence 0.5 [0.800 MR^2 x 566.44  + M x 196.51] = 3502 + MgH

or 394H + 3502 =  [0.400 (40.1631)0.589^2 x 566.44  + 40.1631 x 98.255]

or 394H + 3502 = 3156.979 + 3946.2254

Hence 394 H = 3601.2044

Therefore the value of H, upto which the wheel will reach is 9.14 Metres