Questions 2) Repeat the calculation of part B, this time with the string wrapped

Question

Questions

2)
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 = 2.57cm and mass M1 = 0.790kg and the other with radius R2 = 5.02cm and mass M2 = 1.64kg, are welded together and mounted on a frictionless axis through their common center. 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.

Explanation / Answer

This problem can be done by equating the initial potential energy to the final Translational and rotational kinetic energies.

initial potential energy = m*g*h

final kinetic energy:

0.5I1w^2+0.5I2w^2+0.5mv^2 (i1 = moment of inertia of first disk,i2 = moment of inertia of second disk)

since the string is tied to the smaller disk, v =wr => v= w*2.57*10^-2

applying the formula

1,5*9.8*1.97 = (0.5*0.5*0.79*(0.0257)^2 + 0.5*0.5*1.64*(0.05)^2 + 0.5*1.5*(0.0257)^2) *w^2

w^2 = 17318

w = 131.6 rad/s

v = 131.6*0.0257 = 3.38m/s

b) with the string wrapped around the larger disk.

equation is

1,5*9.8*1.97 = (0.5*0.5*0.79*(0.0257)^2 + 0.5*0.5*1.64*(0.05)^2 + 0.5*1.5*(0.05)^2) *w^2

w= 97.137 rad/s

v =97.137*0.05 = 4.85 m/s

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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