Work and Rotational Kinetic Energy Problems and Solutions

 Problem #1

A rope is wrapped abound a cylindrical drum as shown below. It is pulled with a constant tension of 100 N for six revolutions of the drum. The drum has a radius of 0.500 m. A brake is also applying a force to the drum. The brake pushes inwards on the drum with a force of 200 N. The pressure point is 0.350 m from the centre of the drum. The coefficient of kinetic friction between the brake and the drum is 0.50. Determine the work done by each torque.

Answer:
Work is defined by the formula W = τΔφ in rotational cases. Since the rope does not slip as it is pulled, the object rotates 6 times clockwise or Δφ = ­12π. We know f_k = μ_kN. In this problem, N equals how hard the brake is pressed. Note that the tension and the friction are tangential to the drum.
(a) W_T = (­RT)Δφ = ­(0.5m)(100 N)(­12π) = 1.88 × 10³ J.
(b) W_f = (rf_k)Δφ = (0.35 m)(0.50 × 200 N)(­12π) = ­1.32 × 10³ J.

Problem#2
A rope is wrapped exactly three times around a cylinder with a fixed axis of rotation at its centre. The
cylinder has a mass of 250 kg and a diameter of 34.0 cm. The rope is pulled with a constant tension of 12.6 N. The moment of inertia of a cylinder about its centre is I = ½MR². (a) What is the work down by the rope as it is pulled off the cylinder. Note that the rope does not slip. (b) If the cylinder was initially at rest, what is its final angular velocity? Note that ropes are always tangential to the surfaces that they are wrapped around. Note that the work done by the tension is non­conservative.

Answer:
Since the problem involves a change is speed, we make use of the Generalized Work­Energy Theorem.

Since there is only a change in rotational speed,
W_NC = ΔE = K_f –­ K_i = ½I[(ω_f)² –­ (ω₀)²] = ½I(ω_f)²
The nonconservative force in this problem is tension, W_NC = W_T. So we have
W_T = ½I(ω_f)²                       (1)

(a) The definition of work in rotational situations is W = τΔφ. Tension is always tangential to cylinders so τT = –­RT. Since the rope does not slip, the cylinder rotates clockwise three times so Δφ = ­–6π. We can thus find the work done by the tension
W_T = (–­RT)( –­6π) = (0.34 m)(12.6 N)6π = 40.4 J

(b) Then we find the final velocity from equation (1),

ω_f = [2W_T/I]^½ .

According to the table of Moments of Inertia, I = ½MR² for a solid cylinder. So

ω_f = [4W_T/MR²]^½ = 6.10 rad/s

Problem #3
A large cylinder of mass M = 150 kg and radius R = 0.350 m. The axle on which the cylinder rotates is NOT frictionless. A rope is wrapped around the cylinder exactly ten times. From rest, the rope is pulled with a constant tension of 25.0 N. The rope does not slip and when the rope comes free, the cylinder has a forward angular velocity ω_f = 10.5 rad/s. The moment of inertia of a cylinder is I = ½MR².
(a) What angle was the cylinder rotated through?
(b) What is the frictional torque of the axle?
(c) How long will it take the frictional torque to bring the cylinder to a stop?
(d) How many revolutions will it have turned?

Answer:
(a) Since the rope does not slip, the cylinder rotates ten times so Δφ = ­20π, where the rotation is assumed to be clockwise.

(b) Since the problem involves forces and a change is rotational speed, we make use of the Generalized

Work-­Energy Theorem. Since there is only a change in rotational kinetic energy,

W_NC = ΔE = K_f –­ K_i = ½I[(ω_f)² –­ (ω⁰)²] = ½I(ω_f)²

The nonconservative forces in this problem are the tension and the axle friction, W_NC = W_T + W_f. So we have

W_T + W_f = ½I(ω_f)²

The definition of work in rotational situations is W = τΔφ. Tension is always tangential to cylinders so τT = –­RT, again assuming the rope pulls the cylinder clockwise. Thus the work done by the tension is

W_T = (–­RT)(–­20π) = (0.35 m)(25 N)(20) = 549.78 J

Using this result and equation (1), we can find the work done by friction

W_f = ½I(ω_f)² –­ W_T = ¼M(R_f)² –­ W_T = 506.46 J – 549.78 J = ­43.3178 J

Since W_f = τ_fΔφ, we find the frictional torque to be

τ_f = W_f/Δφ = W_f/­20π = +0.6894 Nm ;

the sign indicating that it is counterclockwise

(c) After the rope comes off the cylinder, the only force acting is friction so the generalized Work­-Energy Theorem becomes

W_f = ΔE = K_f –­ K_i = ½I[(ω_f)² – (ω₀)²] = ­½I(ω₀)²

where ω₀ here is ωf from the first part of the problem and the new ω_f = 0 since the drum comes to a stop. Again W_f = τ_fΔφ, where τ_f is the result from part (b). Therefore

τ_fΔφ = ­½I(ω₀)² = ­(506.46 J)/0.6894 Nm = ­734.6 rad = 117 rev clockwise.

(d) To find the time it takes to slow down, note that we have the initialand finalangular velocities and the angular displacement. Referring to our kinematics equations, we find

Δφ = ½(ω_f + ω₀)t.

Rearranging for t yields,

t = 2Δφ/(ω_f + ω₀) = 140 s