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

Answer:

The value is  $h = 32.91 \ m$

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

      $r = \frac{d}{2}$

=>     $r = \frac{0.52}{2}$

=>    $r = 0.26 \ m$

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

=>  $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} Iw^2 + \frac{1}{2} Iw^2$

=>    $Mgh = \frac{1}{2} * Mv^2 + Iw^2$

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

     $w = \frac{v }{r }$

So

    $Mgh = \frac{1}{2} * Mv^2 + I \frac{v}{r} ^2$

Here $M = m_r + m$

         $M = 320 + 75$

          $M = 395 \ kg$

$395 * 9.8 * h = 0.5 * 395 * (23.61)^2 + 2.1 *[\frac{ 23.61}{ 0.26} ] ^2$

=>   $h = 32.91 \ m$