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

The value is
`P = 14.7 \ N`

Step-by-step 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 = (W)/(g )`

=>
`m = ( 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 = (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 = (v)/(r)`

=>
`w_f = ( 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* (P)/(4.081 *0.25 ) * 20`

=>
`24^2 = 0^2 + 2* (P)/(mr) * 20`

=>
`P = 14.7 \ N`