Question

A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 40 N and a radius of 0.25m that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force P for a distance of 5 m, at which point the end of the rope is moving 6 m/s. If the rope does not slip on the cylinder, what is the value of P

Answer 1

Answer:

The value is $P = 14.7 \ N$

Explanation:

From the question we are told that

    The weight of the hollow cylinder is  $W = 40 \ N$

     The radius of the hollow cylinder is  $r = 0.25 \ m$

       The distance which it is pulled is  $d = 5 \ m$

       The velocity of the end of the rope  is  $v = 6 \ m/s$

Gnerally the mass of the hollow cylinder is  

      $m = \frac{W}{g }$

=>    $m = \frac{ 40 }{ 9.8 }$

=>    $m = 4.081 \ kg$

Generally angular displacement for the distance covered is mathematically represented as

        $\theta = 2 \pi * \frac{ d } {2\pi r }$

=>     $\theta = 2 \pi * \frac{ 5 } {2\pi r }$

=>     $\theta = \frac{ 5 } { 0.25}$

=>   $\theta =20$

Generally the torque experienced  by the hollow cylinder is mathematically represented as

     $P * r = I * \alpha$

Here I  is the moment of inertia

=>   $P * r = m r^2 * \alpha$

=>    $\alpha = \frac{P }{ mr }$

Generally from kinematic equation

     $w_f ^2 = w_i ^2 + 2\alpha \theta$

=>  $w_f ^2 = w_i ^2 + 2\alpha \theta$  

Generally the  final angular velocity is mathematically

      $w_f = \frac{v}{r}$

=>    $w_f = \frac{ 6 }{ 0.25 }$

=>    $w_f = 24 \ m/s$

Generally the  initial  angular velocity is Zero given that the hollow cylinder was at rest before rolling

     $24^2 = 0^2 + 2* \frac{P}{4.081 *0.25 } * 20$

=>   $24^2 = 0^2 + 2* \frac{P}{mr} * 20$

=>   $P = 14.7 \ N$