Answer:
The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex]. (T₁ : T₂) = 1 : 4
Explanation:
Let suppose that both objects are moving along the circular paths at constant speed, such that period of rotation of each object is represented by the following formula:
[tex]\omega = \frac{2\pi}{T}[/tex]
Where:
[tex]\omega[/tex] - Angular speed, measured in radians per second.
[tex]T[/tex] - Period, measured in seconds.
The period is now cleared:
[tex]T = \frac{2\pi}{\omega}[/tex]
Angular speed ([tex]\omega[/tex]) and linear speed ([tex]v[/tex]) are related to each other by this formula:
[tex]\omega = \frac{v}{R}[/tex]
Where [tex]R[/tex] is the radius of rotation, measured in meters.
The angular speed can be replaced and the resultant expression is obtained:
[tex]T = \frac{2\pi\cdot R}{v}[/tex]
Which means that time period is directly proportional to linear speed and directly proportional to radius of rotation. Then, the following relationship is constructed and described below:
[tex]\frac{T_{2}}{T_{1}} = \left(\frac{v_{1}}{v_{2}}\right)\cdot \left(\frac{R_{2}}{R_{1}} \right)[/tex]
Where:
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Time periods of objects 1 and 2, measured in seconds.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Linear speed of objects 1 and 2, measured in meters per second.
[tex]R_{1}[/tex], [tex]R_{2}[/tex] - Radius of rotation of objects 1 and 2, measured in meters.
If [tex]\frac{v_{1}}{v_{2}} = 2[/tex], [tex]R_{1} = 10\,m[/tex] and [tex]R_{2} = 20\,m[/tex], the ratio of time periods is:
[tex]\frac{T_{2}}{T_{1}} = 2\cdot \left(\frac{20\,m}{10\,m} \right)[/tex]
[tex]\frac{T_{2}}{T_{1}} = 4[/tex]
[tex]\frac{T_{1}}{T_{2}} = \frac{1}{4}[/tex]
The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex].
