A small disk of radius R₁ is fastened coaxially to a larger disk of radius R₂. The combination is free to rotate on a fixed axle, which is perpendicular to a horizontal frictionless table top, as shown in the overhead veiw below. The rotational inertia of the combination is I. A string is wrapped around the larger disk and attached to a block of mass m, on the table. Another string is wrapped around the smaller disk and is pulled with a force as shown. The tension in the string pulling the block is: A) R₁F/R₂ B) mR₁R₂F/(I - mR²₂) C) mR₁R₂F/(I + mR²₂) D) mR₁R₂F/(I - mR₁R ₂) E) mR₁R₂F/(I + mR₁R ₂)

Question

Quizplus · Accepted Answer

We can use the principle of conservation of energy to solve this problem. The work done by the tension force on the smaller disk is equal to the change in kinetic energy of the combined system. Let the angular velocities of the smaller and larger disks be ω1 and ω2 respectively. Then the kinetic energy of the system is: K = (1/2)I(ω1^2 + ω2^2) The work done by the tension force on the smaller disk is: W = Fd = (11eaafba_af41_9513_a7a5_3716b2c74118_TB3976_00)R₁ where d is the distance moved by the smaller disk. Since the two disks are connected by a fixed axle, they have the same angular velocity, i.e. ω1 = ω2 = ω. Therefore: K = (1/2)I(2ω^2) = Iω^2 And: W = Fd = (11eaafba_af41_9513_a7a5_3716b2c74118_TB3976_00)R₁ The work done by the tension force is also equal to the work done on the block by the gravitational force, i.e.: W = mgh where g is the acceleration due to gravity and h is the distance the block moves vertically. Since the table top is frictionless, the block does not move horizontally and the distance moved by the smaller disk is equal to the distance moved by the block, i.e. d = hR₂. Therefore: mgh = (11eaafba_af41_9513_a7a5_3716b2c74118_TB3976_00)R₁R₂ Solving for the tension force F: F = (mR₁R₂)/(I + mR₁R ₂) So the answer is C.

A Small Disk of Radius R1 Is Fastened Coaxially to a Larger

A Small Disk of Radius R₁ Is Fastened Coaxially to a Larger