Question

a yo yo has aspool of mass m and radius R. a massless string is wound around an axle of radius b and of negligible mass.if the yo yo released from rest has a downward acceleration of g/9 what is the ratio of R/b?

Answer 1

The ratio of the spool's radius to the axle's radius in a yo-yo, where the yo-yo accelerates downward at g/9 when released, is 3:1. This is found by relating the linear and angular acceleration through Newton's second law and the rotational analogue.

Step-by-step explanation:

The question involves finding the relationship between the spool's radius (R) and the axle's radius (b) of a yo-yo given that the yo-yo accelerates downward at g/9 when released from rest. To solve this, we can use Newton's second law and the rotational analogue to relate the linear acceleration to the angular acceleration and then find the ratio R/b.

Let's assume the tension in the string is T and the moment of inertia of the spool is I. Using Newton's second law for rotation, Tb = Iα, where α is the angular acceleration. The moment of inertia for a solid cylinder (spool) is I = 1/2 mR^2. Therefore, the equation becomes Tb = (1/2 mR^2)(α). For the linear motion, mg - T = ma where a = g/9.

By substituting a = Rα into the linear motion equation and solving the system of equations, we can find that the ratio R/b equals 3. Therefore, the ratio of the spool's radius to the axle's radius is 3:1.