Question

A horizontal disc rotates with a constant angular velocity ω about a vertical axis passing through its centre. A small body m moves along a diameter with a velocity v. Find the force the disc exerts on the body when it is at a distance r located from the rotation axis.

• A. mrω² + 2mvω
• B. mg + $\sqrt { m ^ { 2 } r ^ { 2 } \omega ^ { 4 } + ( 2 m v \omega ) ^ { 2 } }$
• C. $\sqrt { \mathrm { m } ^ { 2 } \mathrm {~g} ^ { 2 } + ( 2 \mathrm { mv } \omega ) ^ { 2 } } + \mathrm { mr } \omega ^ { 2 }$
• D. $m \sqrt { g ^ { 2 } + r ^ { 2 } \omega ^ { 4 } + ( 2 v \omega ) ^ { 2 } }$

Answer 1

Answer: (D)

$m \sqrt { g ^ { 2 } + r ^ { 2 } \omega ^ { 4 } + ( 2 v \omega ) ^ { 2 } }$

Answer 2

The force mg is vertical, 2mvω perpendicular to vertical plane and mrω² outward along the diameter. The resultant force is

F = $m \sqrt { g ^ { 2 } + r ^ { 2 } \omega ^ { 4 } + ( 2 v \omega ) ^ { 2 } }$