beneathstillwaters

A yo-yo is made of two solid cylindrical disks, each of mass 4.6?10^(-2) kg and diameter 7.0?10^(-2) m, joined by a (concentric) thin solid cylindrical hub of mass 7.0?10(-3) kg and diameter 1.1?10(-2) m.

Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its 1.5m long string, if it is released from rest.

What fraction of its kinetic energy is rotational? (K_rotational / K_total)

We assume the tangential velocity of the small cylinder (the spool) is v = wr; where r = 1.1E-2 m. Assuming no slippage between the string and the spool, the hub’s velocity is also v.

From the conservation of energy the potential energy for the spool with mass m = 7E-3 kg and each disc M = 4.6E-2 kg is PE = (m + 2M)gH = 2 1/2 kMR^2 w^2 + 1/2 Kmr^2 w^2 + 1/2 (m + 2M)v^2 = KE which is the kinetic energy near the bottom of the drop H = 1.5 m As w = v/r for the discs and the spool we can make the substitutions. R = 1/2 7E-2 m and r = 1/2 1.1E-2 m.

(m + 2M)gH = kM(R/r)^2 v^2 + 1/2 K m v^2 + 1/2 (m + 2M) v^2 = 1/2 v^2 (kM(R/r)^2 + Km) + 1/2 (m + 2M) v^2 = 1/2 v^2 [ kM(R/r)^2 + Km + m + 2M] = 1/2 v^2 [M( k(R/r)^2 + 2) + m(K + 1)]

Solve for v^2 = 2[(m + 2M)gH]/[M( k(R/r)^2 + 2) + m(K + 1)] and v = sqrt(v^2).

You need to look up k and K for the two solid discs and one cylindrical spool. H = 1.5 meters, the drop distance. Then everything on the RHS of the above equation is given; plug and chug.

[NB: Check my work... the physics is correct, but I may have misplaced a variable or two. I did a units analysis and the RHS does give (m/sec)^2 units for v^2, which is encouraging.]

oldprof | Jul 25, 2009