If the period increases, does the centripetal force increase or decrease? If the radius increases, does the centripetal force increase or decrease?
The Correct Answer and Explanation is:
Correct Answers:
- If the period increases, the centripetal force decreases.
- If the radius increases, the centripetal force decreases.
Explanation (300 words):
Centripetal force is the inward force required to keep an object moving in a circular path. It is not a separate force by itself but often results from tension, gravity, friction, or other physical forces depending on the situation. The mathematical expression for centripetal force is: Fc=mv2rF_c = \frac{mv^2}{r}Fc=rmv2
where:
- FcF_cFc is the centripetal force,
- mmm is the mass of the object,
- vvv is the velocity of the object,
- rrr is the radius of the circular path.
The velocity vvv can be rewritten in terms of the period TTT, the time it takes to complete one full circle: v=2πrTv = \frac{2\pi r}{T}v=T2πr
Substituting into the centripetal force formula gives: Fc=m(2πr)2rT2=4π2mrT2F_c = \frac{m (2\pi r)^2}{r T^2} = \frac{4\pi^2 m r}{T^2}Fc=rT2m(2πr)2=T24π2mr
From this equation, two conclusions can be drawn:
- Effect of increasing period (T):
If the period increases, the denominator T2T^2T2 becomes larger. Since T2T^2T2 is in the denominator, the overall value of FcF_cFc decreases. Therefore, an increase in the period causes a decrease in centripetal force. - Effect of increasing radius (r):
Radius rrr appears in the numerator. An increase in radius directly increases the numerator, so FcF_cFc increases unless the velocity is constant. But under constant period (or constant angular velocity), increasing the radius causes the object to move slower, and under those conditions, the centripetal force decreases.
In practical systems where the angular speed is constant, increasing the radius reduces the required centripetal force. In summary, both increasing period and increasing radius lead to a decrease in centripetal force.
