Question

A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 40.0 N and a radius of 0.25 m, 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 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is the value of P?

Answer 1

Answer:

14.67889 N

Explanation:

M = Mass of cylinder = $\dfrac{40}{9.81}\ kg$

R = Radius = 0.25 m

d = Distance = 5 m

v = Velocity = 6 m/s

As there are no other energy in the system only the kinetic energy and the excess work is present

$K_i+U_i+W_o=K_f+U_f\\\Rightarrow W_o=K_f$

Work done by applied forces is

$W_o=Pd$

Moment of inertia is given by

$I=MR^2$

Angular speed is given by

$\omega=\dfrac{v}{R}$

The kinetic energy is

$K_f=\dfrac{1}{2}I\omega^2$

So,

$Pd=\dfrac{1}{2}I\omega^2\\\Rightarrow P=\dfrac{1}{2d}MR^2\dfrac{v^2}{R^2}\\\Rightarrow P=\dfrac{Mv^2}{2d}\\\Rightarrow P=\dfrac{\dfrac{40}{9.81}\times 6^2}{2\times 5}\\\Rightarrow P=14.67889\ N$

The force is 14.67889 N