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Deffense [45]
3 years ago
6

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 rotate

s 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
Physics
1 answer:
leonid [27]3 years ago
7 0

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

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