Two metal disks, one with radius R1= 2.36cm and mass M1= 0.78 kg and the other w

Question

Two metal disks, one with radius R1= 2.36cm and mass M1= 0.78 kg and the other with radius R2= 4.95cm and mass M2= 1.66 kg are welded together and mounted on a frictionless axis through their common center.

What is the total moment of inertia of the two disks?

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 1.97m above the floor, what is its speed just before it strikes the floor?

Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.

Two metal disks, one with radius R1= 2.36cm and mass M1= 0.78 kg and the other with radius R2= 4.95cm and mass M2= 1.66 kg are welded together and mounted on a frictionless axis through their common center. What is the total moment of inertia of the two disks? 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 1.97m above the floor, what is its speed just before it strikes the floor?

Explanation / Answer

If the discs have a common axis, then the moments can be directly summed. The moment of a single disc is m*r/2, so the moment of the two discs is 0.8*0.025/2 + 1.60*0.05/2; I = 0.00225 kg-m

Let Ft be the tension in the spring. The force balance on the hanging block (mass mb) is

mb*g - Ft = mb*a

The angular acceleration of the discs is ? = T/I where T = torque on the discs. T = Ft*r so ? = Ft*r/I but a = ?*r a = Ft*r/I. Solve the above eq for Ft and insert; Ft = mb*g = mb*a

a = (mb*g - mb*a)*r/I = g*mb*r/I - mb*a*r/I

a*(1 + mb*r/I) = mb*r/I

a = mb*r/I / (1 + mb*r/I) = 2.88 m/s

h = 0.5*a*t so the time to reach the floor is ?[2*h/a]. The speed is a*t, so v - ?[2*h*a] = 3.4 m/s

Alternate (and simpler) approach:

Initial energy of the system is mb*g*h

Final energy is 0.5*mb*v + 0.5*I*? ? = v/r so

mb*g*h = 0.5*mb*v + 0.5*I*v/r = v*( 0.5*mb + 0.5*I/r)

v = ?[2*mb*g*h/(mb + I/r)]

v = 3.4 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.