Answer:
The value is [tex]P = 14.7 \ N[/tex]
Explanation:
From the question we are told that
The weight of the hollow cylinder is [tex]W = 40 \ N[/tex]
The radius of the hollow cylinder is [tex]r = 0.25 \ m[/tex]
The distance which it is pulled is [tex]d = 5 \ m[/tex]
The velocity of the end of the rope is [tex]v = 6 \ m/s[/tex]
Gnerally the mass of the hollow cylinder is
[tex]m = \frac{W}{g }[/tex]
=> [tex]m = \frac{ 40 }{ 9.8 }[/tex]
=> [tex]m = 4.081 \ kg[/tex]
Generally angular displacement for the distance covered is mathematically represented as
[tex]\theta = 2 \pi * \frac{ d } {2\pi r }[/tex]
=> [tex]\theta = 2 \pi * \frac{ 5 } {2\pi r }[/tex]
=> [tex]\theta = \frac{ 5 } { 0.25}[/tex]
=> [tex]\theta =20[/tex]
Generally the torque experienced by the hollow cylinder is mathematically represented as
[tex]P * r = I * \alpha[/tex]
Here I is the moment of inertia
=> [tex]P * r = m r^2 * \alpha[/tex]
=> [tex]\alpha = \frac{P }{ mr }[/tex]
Generally from kinematic equation
[tex]w_f ^2 = w_i ^2 + 2\alpha \theta[/tex]
=> [tex]w_f ^2 = w_i ^2 + 2\alpha \theta[/tex]
Generally the final angular velocity is mathematically
[tex]w_f = \frac{v}{r}[/tex]
=> [tex]w_f = \frac{ 6 }{ 0.25 }[/tex]
=> [tex]w_f = 24 \ m/s[/tex]
Generally the initial angular velocity is Zero given that the hollow cylinder was at rest before rolling
[tex]24^2 = 0^2 + 2* \frac{P}{4.081 *0.25 } * 20[/tex]
=> [tex]24^2 = 0^2 + 2* \frac{P}{mr} * 20[/tex]
=> [tex]P = 14.7 \ N[/tex]
