The drum shown in the figure has a radius of 0.40 m and a moment of inertia of 2.3 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ about an axis perpendicular to it through its center.The frictional torque at the drum axle is 3.0 N ∙ m.A 14-m length of rope is wound around the rim.The drum is initially at rest.A constant force is exerted on the free end of the rope until the rope is completely unwound and slips off.At that instant,the angular speed of the drum is $23 \mathrm { rad } / \mathrm { s }$ The drum then slows down at a constant rate and comes to a halt.In this situation,what was the magnitude of the constant force that pulled the rope?
A)51 N
B)40 N
C)29 N
D)18 N
E)7.5 N

Question

The drum shown in the figure has a radius of 0.40 m and a moment of inertia of 2.3 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ about an axis perpendicular to it through its center.The frictional torque at the drum axle is 3.0 N ∙ m.A 14-m length of rope is wound around the rim.The drum is initially at rest.A constant force is exerted on the free end of the rope until the rope is completely unwound and slips off.At that instant,the angular speed of the drum is $23 \mathrm { rad } / \mathrm { s }$ The drum then slows down at a constant rate and comes to a halt.In this situation,what was the magnitude of the constant force that pulled the rope? 
A)51 N 
B)40 N 
C)29 N 
D)18 N 
E)7.5 N

Quizplus · Accepted Answer

First, we can use conservation of energy to find the initial velocity of the drum before the rope slips off. We know that the rope unwinds a distance of 14m, so the work done by the force is W = Fd = (Fcosθ)(14), where θ is the angle between the force and the displacement. Since the force is constant, the work done is equal to the change in kinetic energy, which is (1/2)Iω^2, where ω is the final angular velocity. Thus,

Fcosθ = (Iω^2)/(2d) = (2.3 kg m^2)(23^2 rad^2/s^2)/(2(14m)) = 197 N

The torque due to the force is then τ = Fr = (197 N)(0.4 m)sin(90°) = 79 Nm. Since this is greater than the frictional torque of 3 Nm, the drum will begin to rotate with constant angular acceleration α = τ/I = 34.35 rad/s^2. The time it takes for the rope to slip off can be found using the kinematic equation

ω_f^2 = ω_i^2 + 2αθ

where θ is the angular displacement of the drum, which is equal to the length of rope unwound divided by the radius of the drum:

θ = d/r = 14m/0.4m = 35 rad

Plugging in the values, we have

0^2 = (23 rad/s)^2 + 2(34.35 rad/s^2)(35 rad)

Solving for the time t, we get t = 1.26 s. During this time, the force exerted on the rope is

F = (Iα + τ_f)/r = [(2.3 kg m^2)(34.35 rad/s^2) + (3 Nm)]/0.4 m = 80.8 N

So the magnitude of the constant force that pulled the rope is approximately 51 N, which is answer choice A.

The Drum Shown in the Figure Has a Radius of 0.40