Question

Each wheel of a 320 kg motorcycle is 52 cm in diameter and has rotational inertia 2.1 kg m2 . The cycle and its 75 kg rider are coasting at 85 km/h on a flat road when they encounter a hill. If the cycle rolls up the hill with no applied power and no significant internal friction, what vertical height will it reach

Answer 1

The value is
`h = 32.91 \ m`

Step-by-step explanation:

From the question we are told that

The diameter of each wheel is
`d = 52 \ cm = 0.52 \ m`

The mass of the motorcycle is
`m = 320 \ kg`

The rotational kinetic inertia is
`I = 2.1 \ kg \ m^2`

The mass of the rider is
`m_r = 75 \ kg`

The velocity is
`v = 85 \ km/hr = 23.61 \ m/s`

Generally the radius of the wheel is mathematically represented as

`r = (d)/(2)`

=>
`r = (0.52)/(2)`

=>
`r = 0.26 \ m`

Generally from the law of energy conservation

Potential energy attained by system(motorcycle and rider ) = Kinetic energy of the system + rotational kinetic energy of both wheels of the motorcycle

=>
`Mgh = (1)/(2) Mv^2 + (1)/(2) Iw^2 + (1)/(2) Iw^2`

=>
`Mgh = (1)/(2) * Mv^2 + Iw^2`

Here
`w` is the angular velocity which is mathematically represented as

`w = (v )/(r )`

So

`Mgh = (1)/(2) * Mv^2 + I (v)/(r) ^2`

Here
`M = m_r + m`

`M = 320 + 75`

`M = 395 \ kg`

`395 * 9.8 * h = 0.5 * 395 * (23.61)^2 + 2.1 *[( 23.61)/( 0.26) ] ^2`

=>
`h = 32.91 \ m`