Question

In circular motion, what are the possible values (zero, positive or negative) of the following: (a) omega.v (b) v.a, (c) omega.alpha

Answer 1

1. omega.v will always be 0
2. v.a can be positive, negative or 0
3. omega.alpha can be positive, negative or 0.

1. omega.v

We know that linear velocity vector is cross product of omega vector and radius. Thus omega vector is perpendicular to plane of linear velocity and radius vector.

ω.v= |w| * |v| * cos Ф

Ф= 90.

Hence omega.v=0.

2. v.a

For uniform motion, v and a are perpendicular. Hence Ф is 90 and dot product becomes zero.

For positive accceleration, Ф will be acute,cos Ф will be positive and hence, v.a is positive.

For negative acceleration, Ф will be obtuse, cosФ will be negative and hence, v.a will be negative.

3. omega.alpha

Alpha vector is perpendicular to plane of liner acceleration and radius vector. Omega vector is also perpendicular to velocity and radius vector.

For uniform motion, acceleration is 0 and hence angular acceleration alpha is zero. Omega.alpha will be zero.

Omega and angular acceleration can be anti parallel. Thus angle between them can be 180. Thus omega.alpha will be negative since cos 180 is negative.

Omega and acceleration can be parallel. Thus angle between them can be zero.Thus omega.alpha will be positive since cos 0 is positive.