10R1R R2+12R2 5. One way to determine the moment of inertia of a disk of radius R and mass M in the laboratory is to suspend a mass m from it to give the disk an angular acceleration. The mass m is attached to a string that is wound around a small hub of radius r as shown in Figure 7. The mass and moment of the inertia of the hub can be neglected. The mass m is released and takes a time t to fall Fig. 7 distance h to the floor. The total torque on the disk is due to the frictional torque t frictm of the axle on the disk. Once the mass m hits the floor, the disk slowly stops spinning during an additional time t 'because of the sole influence of the small frictional torque on the disk. Find the moment of inertia of the disk in terms of m, r, g, h, t and t Solution Assume the acceleration of the falling mass is a, and the angular acceleration of the disk is a. The angular speed of the disk is @=at while the falling mass reaches the floor. So 2h h Using newton nd law T=ma Tr-t.=la Apply the rotational analog of Newton's second law to the disk △t During the course of the mass hits the floor Using the equation(1)and (2), we get the moment of inertia of the disk 1=m(8-a)12 2 1 2 1 2 1 0 1 2 2 I R I R I R R + = ω ω 5. One way to determine the moment of inertia of a disk of radius R and mass M in the laboratory is to suspend a mass m from it to give the disk an angular acceleration. The mass m is attached to a string that is wound around a small hub of radius r, as shown in Figure 7. The mass and moment of the inertia of the hub can be neglected. The mass m is released and takes a time t to fall a distance h to the floor. The total torque on the disk is due to the torque of the tension in the string and the small but unknown frictional torque frictn τ r of the axle on the disk. Once the mass m hits the floor, the disk slowly stops spinning during an additional time t ′ because of the sole influence of the small frictional torque on the disk. Find the moment of inertia of the disk in terms of m, r, g, h, t and t ′. Solution: Assume the acceleration of the falling mass is a, and the angular acceleration of the disk isα . The angular speed of the disk is ω = αt while the falling mass reaches the floor. So 2 2 2 2 1 t h h = at ⇒ a = Using Newton’s second law ⎪ ⎩ ⎪ ⎨ ⎧ = − = − = α τ α a r Tr I mg T ma fric (1) Apply the rotational analog of Newton’s second law to the disk t Ltotal L total ∆ ∆ = ≈ r r r dt d τ During the course of the mass hits the floor ' ' 0 t It I t fric ω α τ = − = (2) Using the equation (1) and (2), we get the moment of inertia of the disk ' 1 1) 2 ( ' 1 ( ) 1 2 2 2 t t h gt mr t a t mr g a I + − = + ⋅ − = m M R Fig.7 r Wheel Hub