Advertisements
Advertisements
Question
Obtain expressions of energy of a particle at different positions in the vertical circular motion .
Solution
Lowest point of circular motion:
Total energy = KE
\[ = \frac{1}{2} {mv}^2 \]
Substitute v = \[\sqrt{5rg}\]
Total energy =\[ \frac{1}{2}m \left[ \sqrt{5rg} \right]^2 \]
\[ = \frac{5mrg}{2}\]
Highest point of circular motion:
Total enery = PE + KE
= mgh + KE
\[ = mg\left[ 2r \right] + \frac{1}{2} {mv}^2 \]
Substitute v =\[ \sqrt{rg}\]
Total energy = 2mgr \[+ \frac{1}{2}m \left[ \sqrt{rg} \right]^2 \]
\[ = 2mgr + \frac{mgr}{2}\]
\[ = \frac{5mrg}{2}\]
APPEARS IN
RELATED QUESTIONS
The difference in tensions in the string at lowest and highest points in the path of the particle of mass 'm' performing vertical circular motion is:....
a) 2 mg
b) 4 mg
c) 6 mg
d) 8 mg
A planet is revolving around a star in a circular orbit of radius R with a period T. If the
gravitational force between the planet and the star is proportional to `R^(-3/2)` then
A) `T^2 prop R^(5/2)`
B) `T^2 prop R^((-7)/2)`
C) `T^2 prop R^(3/2)`
D) `T^2 prop R^4`
A particle of mass m performs the vertical motion in a circle of radius r. Its potential energy at the highest point is _______. (g is acceleration due to gravity)
