Question

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center. What is the ratio of the tangential acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod

Answer 1

Answer:

Explanation:

angular acceleration ω = 8 rad /s²

tangential acceleration a = angular acceleration x radius of circle of rotation

For a point on the end of the rod , radius of circle = L .

tangential acceleration = 8 L .

For  a point a distance L/2 from the end of the rod, radius of circle = L /2 .

tangential acceleration = 8 L/2 = 4 L  .

It is so because angular acceleration will be same for this point , only radius changes .

required ratio = 8L / 4L

= 2.