Question

g I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.6 kg and radius r=0.37 m from rest at an initial height of h=6.7 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.7 m/s. What is the rotational inertia of my wheel?

Answer 1

The rotational inertia of my wheel is
`I =1.083 \ kg \cdot m^2`

Step-by-step explanation:

From the question we are told that

The mass of the wheel is
`m = 1.6 \ kg`

The radius of the wheel is
`r = 0.37 \ m`

The height is
`h = 6.7 m`

The linear speed is
`v = 4.7 m/s`

According to the law of energy conservation

`PE = KE + KE_R`

Where PE is the potential energy at the height h which is mathematically represented as

`PE = mgh`

While KE is the kinetic energy at the bottom of height h

`KE = (1)/(2) mv^2`

Where
`KE_R` is the rotational kinetic energy which is mathematically represented as

`KE_R = (1)/(2) * I * (v^2)/(r^2)`

Where
`I` is the rotational inertia

So substituting this formula into the equation of energy conservation

`mgh = (1)/(2) mv^2 + (1)/(2) * I * (v^2)/(r^2)`

=>
`I =[ \ mgh - (1)/(2) mv^2 \ ]* (2 r^2)/(v^2)`

substituting values

`I =[ \ 1.6 * 9.8 * 6.7 - (1)/(2) * 1.6 *4.7^2 \ ]* (2 * 0.37^2)/(4.7^2)`

`I =1.083 \ kg \cdot m^2`