Two metal disks, one with radius R_1 = 2.56cm and mass M_1= 0.820cm and the othe

Question

Two metal disks, one with radius R_1 = 2.56cm and mass M_1= 0.820cm and the other with radius R_2 = 5.02 and mass M_2= 1.64 , are welded together and mounted on a frictionless axis through their common center. (Figure 1) . A)What is the total moment of inertia of the two disks? B)A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 1.99 above the floor, what is its speed just before it strikes the floor? C)Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.

Explanation / Answer

To find combined rotational inertia, add up individual rotational inertias. Treat them as solid uniform discs. Inet = 1/2*(m1*r1^2 + m2*r2^2) To solve the falling/hanging block, consider forces on hanging block: Weight (M*g): downward Tension (T): upward Newton's 2nd law for the hanging block (+a defined downward): M*g - T = M*a Forces on the combination of discs: Weight ((m1 + m2)*g): downward Bearing reaction force (B): upward Tension (T): downward We aren't interested in balancing forces, because B will be as large as necessary to constrain the axle in place. Instead we are interested in torques. The only force causing a torque is T. T*r1 = Inet*alpha No-slip condition: alpha = a/r1 Summary of equations: T*r1^2 = Inet*a M*g - T = M*a Solve two equations, two unknowns (we are actually only interested in a): T = Inet*a/r1^2 M*g - Inet*a/r1^2 = M*a Gather and factor: M*g = a*(M + Inet/r1^2) Thus: a = g*M*r1^2/(M*r1^2 + Inet) Kinematics equation in which we don't care about time: vf^2 = vi^2 + 2*a*d vi = 0, thus: vf^2 = 2*a*d So, vf = sqrt(2*a*d) Substitute: vf = sqrt(2*d*g*M*r1^2/(M*r1^2 + Inet)) Substitute expression for Inet: vf = sqrt(2*d*g*M*r1^2/(1/2*(m1*r1^2 + m2*r2^2) + M*r1^2 ) ) Simplify: vf = 2*sqrt(d*g*M*r1^2/(m1*r1^2 + m2*r2^2 + 2*M*r1^2) ) For part C, all we do is swap the radius associated with the spooling to r2: vf = 2*sqrt(d*g*M*r2^2/(m1*r1^2 + m2*r2^2 + 2*M*r2^2) ) Summary: A) Inet = 1/2*(m1*r1^2 + m2*r2^2) B) vf = 2*sqrt(d*g*M*r1^2/(m1*r1^2 + m2*r2^2 + 2*M*r1^2) ) C) vf = 2*sqrt(d*g*M*r2^2/(m1*r1^2 + m2*r2^2 + 2*M*r2^2) ) Data: m1:=0.820 kg; m2:=1.64 kg; r1:=0.0256 m; r2:=0.0502 m; M:=1.5 kg; d:=1.99 m; g:=9.8 N/kg; Results: A) Inet = ....kg-m^2 B) vf =....m/s C).....m/s

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